同济大学数学系主编, 高等数学 . 第二版, 下册. 2009年, 同济大学出版社.

7 空间解析几何与向量代数

7.5 空间直线及其方程

1(3). 求过点 P(2,-3,3)<script id="MathJax-Element-1" type="math/tex">P(2,-3,3)</script> 且与平面 \pi: x+2y-3z-2=0<script id="MathJax-Element-2" type="math/tex">\pi: x+2y-3z-2=0</script> 垂直的直线 l<script type="math/tex">l</script> 的方程.

解答: 直线 l<script type="math/tex">l</script> 过点 P(2,-3,3)<script type="math/tex">P(2,-3,3)</script>, 且方向向量与平面法向量 {\bf n}=\sed{1,2,-3}<script type="math/tex">{\bf n}=\sed{1,2,-3}</script> 平行, 为 {\bf s}=\sed{1,2,-3}<script type="math/tex">{\bf s}=\sed{1,2,-3}</script>. 故其方程为 \bex \cfrac{x-2}{1}=\cfrac{y+8}{2}=\cfrac{z-3}{-3}. \eex<script type="math/tex; mode=display">\bex \cfrac{x-2}{1}=\cfrac{y+8}{2}=\cfrac{z-3}{-3}. \eex</script>

4. 求 k<script type="math/tex">k</script> 的值, 使直线 \dps{\cfrac{x-3}{2k} =\cfrac{y+1}{k+1}=\cfrac{z-3}{5}}<script type="math/tex">\dps{\cfrac{x-3}{2k} =\cfrac{y+1}{k+1}=\cfrac{z-3}{5}}</script> 与直线 \dps{\cfrac{x-1}{3}=y+5=\cfrac{z+2}{k-2}}<script type="math/tex">\dps{\cfrac{x-1}{3}=y+5=\cfrac{z+2}{k-2}}</script> 垂直.

解答: 注意到两直线垂直当且仅当它们的方向向量垂直, 于是 \bex 0=\sed{2k,k+1,5}\cdot\sed{3,1,k-2} =6k+(k+1)+5(k-2) =12k-9, \eex<script type="math/tex; mode=display">\bex 0=\sed{2k,k+1,5}\cdot\sed{3,1,k-2} =6k+(k+1)+5(k-2) =12k-9, \eex</script> \bex k=\cfrac{3}{4}. \eex<script type="math/tex; mode=display">\bex k=\cfrac{3}{4}. \eex</script>

6(2). 求直线 \dps{l_1:\ \left\{\ba{ll} 5x-3y+3z-9=0\\ 3x-2y+z-1=0 \ea\right.}<script type="math/tex">\dps{l_1:\ \left\{\ba{ll} 5x-3y+3z-9=0\\ 3x-2y+z-1=0 \ea\right.}</script> 与直线 \dps{l_2:\ \cfrac{x-1}{2}=\cfrac{y+2}{-1}=\cfrac{z}{2}}<script type="math/tex">\dps{l_2:\ \cfrac{x-1}{2}=\cfrac{y+2}{-1}=\cfrac{z}{2}}</script> 的夹角 \varphi<script type="math/tex">\varphi</script>.

解答: 直线 l_1<script type="math/tex">l_1</script> 的方向向量为 \bex {\bf s}_1&=&\sed{5,-3,3}\times \sed{3,-2,1} =\sev{\ba{ccc} {\bf i}&{\bf j}&{\bf k}\\ 5&-3&3\\ 3&-2&1 \ea}\\ &=&\sed{\sev{\ba{cc} -3&3\\ -2&1 \ea},\sev{\ba{cc} 3&5\\ 1&3 \ea},\sev{\ba{cc} 5&-3\\ 3&-2 \ea}}\\ &=&\sed{3,4,-1}; \eex<script type="math/tex; mode=display">\bex {\bf s}_1&=&\sed{5,-3,3}\times \sed{3,-2,1} =\sev{\ba{ccc} {\bf i}&{\bf j}&{\bf k}\\ 5&-3&3\\ 3&-2&1 \ea}\\ &=&\sed{\sev{\ba{cc} -3&3\\ -2&1 \ea},\sev{\ba{cc} 3&5\\ 1&3 \ea},\sev{\ba{cc} 5&-3\\ 3&-2 \ea}}\\ &=&\sed{3,4,-1}; \eex</script> 直线 l_2<script type="math/tex">l_2</script> 的方向向量为 {\bf s}_2=\sed{2,-1,2}<script type="math/tex">{\bf s}_2=\sed{2,-1,2}</script>. 故 \bex \cos\varphi =\cfrac{\sev{{\bf s}_1\cdot {\bf s}_2}}{\sev{{\bf s}_1}\cdot \sev{{\bf s}_2}} =\cfrac{3\cdot 2+4\cdot (-1)+(-1)\cdot 2}{\sqrt{3^2+4^2+(-1)^2}\cdot\sqrt{2^2+(-1)^2+2^2}} =0! \eex<script type="math/tex; mode=display">\bex \cos\varphi =\cfrac{\sev{{\bf s}_1\cdot {\bf s}_2}}{\sev{{\bf s}_1}\cdot \sev{{\bf s}_2}} =\cfrac{3\cdot 2+4\cdot (-1)+(-1)\cdot 2}{\sqrt{3^2+4^2+(-1)^2}\cdot\sqrt{2^2+(-1)^2+2^2}} =0! \eex</script> 于是 \dps{\varphi=\cfrac{\pi}{2}}<script type="math/tex">\dps{\varphi=\cfrac{\pi}{2}}</script>.

7(1). 求直线 \dps{l:\ \cfrac{x-1}{2}=\cfrac{y}{-1}=\cfrac{z+1}{2}}<script type="math/tex">\dps{l:\ \cfrac{x-1}{2}=\cfrac{y}{-1}=\cfrac{z+1}{2}}</script> 与平面 \pi:\ x-y+2z-2=0<script type="math/tex">\pi:\ x-y+2z-2=0</script> 的交点 P<script type="math/tex">P</script> 及夹角 \theta<script type="math/tex">\theta</script>.

解答: 直线 l<script type="math/tex">l</script> 的方向向量为 {\bf s}=\sed{1,1,2}<script type="math/tex">{\bf s}=\sed{1,1,2}</script>; 平面 \pi<script type="math/tex">\pi</script> 的法向量为 {\bf n}=\sed{2,1,1}<script type="math/tex">{\bf n}=\sed{2,1,1}</script>. 于是 \bex \sin\theta =\cfrac{\sev{{\bf s}\cdot{\bf n}}}{\sev{{\bf s}}\cdot\sev{{\bf n}}} =\cfrac{\sev{1\cdot2 +1\cdot 1+2\cdot 1}}{\sqrt{1^2+1^2+2^2}\cdot\sqrt{2^2+1^2+1^2}} =\cfrac{5}{\sqrt{6}\cdot\sqrt{6}} =\cfrac{5}{6}, \eex<script type="math/tex; mode=display">\bex \sin\theta =\cfrac{\sev{{\bf s}\cdot{\bf n}}}{\sev{{\bf s}}\cdot\sev{{\bf n}}} =\cfrac{\sev{1\cdot2 +1\cdot 1+2\cdot 1}}{\sqrt{1^2+1^2+2^2}\cdot\sqrt{2^2+1^2+1^2}} =\cfrac{5}{\sqrt{6}\cdot\sqrt{6}} =\cfrac{5}{6}, \eex</script> \bex \varphi=\arccos\cfrac{5}{6}. \eex<script type="math/tex; mode=display">\bex \varphi=\arccos\cfrac{5}{6}. \eex</script>

7.6旋转曲面与二次曲面

1(1). 求 zOx<script type="math/tex">zOx</script> 平面上的抛物线 z^2=5x<script type="math/tex">z^2=5x</script> 绕 x<script type="math/tex">x</script> 轴旋转一周所成旋转曲面的方程.

解答: 注意到 \bex \mbox{绕 }x\mbox{ 轴旋转 }\ra x\mbox{ 坐标不动, 另一坐标换成正负开根号平方和的形式}, \eex<script type="math/tex; mode=display">\bex \mbox{绕 }x\mbox{ 轴旋转 }\ra x\mbox{ 坐标不动, 另一坐标换成正负开根号平方和的形式}, \eex</script> 于是所求为 \dps{y^2+z^2=\sex{\pm\sqrt{y^2+z^2}}^2=5x}<script type="math/tex">\dps{y^2+z^2=\sex{\pm\sqrt{y^2+z^2}}^2=5x}</script>.

2. 指出下列方程表示什么曲面. 若是旋转曲面 指出它们可以由什么曲线绕什么轴旋转而成.

(2). \dps{x^2+3y^2+2z^2=1}<script type="math/tex">\dps{x^2+3y^2+2z^2=1}</script>;

(5). \dps{x^2-\cfrac{y^2}{2}+z^2=1}<script type="math/tex">\dps{x^2-\cfrac{y^2}{2}+z^2=1}</script>;

(6). \dps{\cfrac{x^2}{4}+\cfrac{y^2}{4}-z^2=-1}<script type="math/tex">\dps{\cfrac{x^2}{4}+\cfrac{y^2}{4}-z^2=-1}</script>.

解答:

(2). 注意到 x^2<script type="math/tex">x^2</script>, y^2<script type="math/tex">y^2</script>, z^2<script type="math/tex">z^2</script> 的系数分别为 1<script type="math/tex">1</script>, 3<script type="math/tex">3</script>, 2<script type="math/tex">2</script>, 互不相同, 而不是旋转曲面, 但确是椭球面.

(5). 这是一张单叶双曲面, 因为其方程右端为 1<script type="math/tex">1</script> 而左端仅有一个负号. 再注意到 x^2<script type="math/tex">x^2</script>, z^2<script type="math/tex">z^2</script> 的系数均为 1<script type="math/tex">1</script>, 而该曲面为旋转曲面, 剩下的 y<script type="math/tex">y</script> 轴就是旋转轴, 而原曲线方程就是在曲面方程中把含 x<script type="math/tex">x</script> 或 z<script type="math/tex">z</script> 的单项式拿掉后的方程. 比如 \bex \ba{cc} \mbox{把 } x\mbox{ 拿掉}:\ -\cfrac{y^2}{2}+z^2=1\mbox{ 绕 }y\mbox{ 轴旋转成 }x^2-\cfrac{y^2}{2}+z^2=1;\\ \mbox{把 } z\mbox{ 拿掉}:\ x^2-\cfrac{y^2}{2}=1\mbox{ 绕 }y\mbox{ 轴旋转成 }x^2-\cfrac{y^2}{2}+z^2=1. \ea \eex<script type="math/tex; mode=display">\bex \ba{cc} \mbox{把 } x\mbox{ 拿掉}:\ -\cfrac{y^2}{2}+z^2=1\mbox{ 绕 }y\mbox{ 轴旋转成 }x^2-\cfrac{y^2}{2}+z^2=1;\\ \mbox{把 } z\mbox{ 拿掉}:\ x^2-\cfrac{y^2}{2}=1\mbox{ 绕 }y\mbox{ 轴旋转成 }x^2-\cfrac{y^2}{2}+z^2=1. \ea \eex</script>

(6). 这是一张双叶双曲面, 因为 \bex \cfrac{x^2}{4}+\cfrac{y^2}{4}-z^2=-1 \ra -\cfrac{x^2}{4}-\cfrac{y^2}{4}+z^=1, \eex<script type="math/tex; mode=display">\bex \cfrac{x^2}{4}+\cfrac{y^2}{4}-z^2=-1 \ra -\cfrac{x^2}{4}-\cfrac{y^2}{4}+z^=1, \eex</script> 右端为 1<script type="math/tex">1</script>, 左端有两个负号. 同

(2) 的过程, z<script type="math/tex">z</script> 为旋转轴, \bex \ba{cc} \mbox{把 } x\mbox{ 拿掉}:\ \cfrac{y^2}{4}-z^2=-1\mbox{ 绕 }z\mbox{ 轴旋转成 }\cfrac{x^2}{4}+\cfrac{y^2}{4}-z^2=-1;\\ \mbox{把 } y\mbox{ 拿掉}:\ \cfrac{x^2}{4}-z^2=-1 \mbox{ 绕 }z\mbox{ 轴旋转成 }\cfrac{x^2}{4}+\cfrac{y^2}{4}-z^2=-1. \ea \eex<script type="math/tex; mode=display">\bex \ba{cc} \mbox{把 } x\mbox{ 拿掉}:\ \cfrac{y^2}{4}-z^2=-1\mbox{ 绕 }z\mbox{ 轴旋转成 }\cfrac{x^2}{4}+\cfrac{y^2}{4}-z^2=-1;\\ \mbox{把 } y\mbox{ 拿掉}:\ \cfrac{x^2}{4}-z^2=-1 \mbox{ 绕 }z\mbox{ 轴旋转成 }\cfrac{x^2}{4}+\cfrac{y^2}{4}-z^2=-1. \ea \eex</script>

5. 分别指出平面 z=2<script type="math/tex">z=2</script>, z-2<script type="math/tex">z-2</script>, x=1<script type="math/tex">x=1</script>, y=1<script type="math/tex">y=1</script> 与双叶双曲面 \dps{-x^2-\cfrac{y^2}{4}+z^2=1}<script type="math/tex">\dps{-x^2-\cfrac{y^2}{4}+z^2=1}</script> 相交 的曲线的类型.

解答: 注意到 \bex & &\left\{\ba{cc} -x^2-\cfrac{y^2}{4}+z^2=1\\ z=2 \ea\right. \ra \left\{\ba{cc} -x^2-\cfrac{y^2}{4}+4=1\\ z=2 \ea\right. \ra \left\{\ba{cc} x^2+\cfrac{y^2}{4}=3\\ z=2 \ea\right.\\ & & \ra \left\{\ba{cc} \cfrac{x^2}{3}+\cfrac{y^2}{12}=1\\ z=2 \ea\right. \mbox{ 是一个椭圆;} \eex<script type="math/tex; mode=display">\bex & &\left\{\ba{cc} -x^2-\cfrac{y^2}{4}+z^2=1\\ z=2 \ea\right. \ra \left\{\ba{cc} -x^2-\cfrac{y^2}{4}+4=1\\ z=2 \ea\right. \ra \left\{\ba{cc} x^2+\cfrac{y^2}{4}=3\\ z=2 \ea\right.\\ & & \ra \left\{\ba{cc} \cfrac{x^2}{3}+\cfrac{y^2}{12}=1\\ z=2 \ea\right. \mbox{ 是一个椭圆;} \eex</script> \bex \left\{\ba{cc} -x^2-\cfrac{y^2}{4}+z^2=1\\ z=-2 \ea\right. \ra \left\{\ba{cc} \cfrac{x^2}{3}+\cfrac{y^2}{12}=1\\ z=2 \ea\right. \mbox{ 是一个椭圆;} \eex<script type="math/tex; mode=display">\bex \left\{\ba{cc} -x^2-\cfrac{y^2}{4}+z^2=1\\ z=-2 \ea\right. \ra \left\{\ba{cc} \cfrac{x^2}{3}+\cfrac{y^2}{12}=1\\ z=2 \ea\right. \mbox{ 是一个椭圆;} \eex</script> \bex \left\{\ba{cc} -x^2-\cfrac{y^2}{4}+z^2=1\\ x=1 \ea\right. \ra \left\{\ba{cc} -\cfrac{y^2}{8}+\cfrac{z^2}{2}=1\\ x=1 \ea\right. \mbox{ 是一条双曲线;} \eex<script type="math/tex; mode=display">\bex \left\{\ba{cc} -x^2-\cfrac{y^2}{4}+z^2=1\\ x=1 \ea\right. \ra \left\{\ba{cc} -\cfrac{y^2}{8}+\cfrac{z^2}{2}=1\\ x=1 \ea\right. \mbox{ 是一条双曲线;} \eex</script> \bex \left\{\ba{cc} -x^2-\cfrac{y^2}{4}+z^2=1\\ y=1 \ea\right. \ra \left\{\ba{cc} -\cfrac{x^2}{5/4}+\cfrac{z^2}{5/4}=1 \\ x=1 \ea\right. \mbox{ 是一条双曲线.} \eex<script type="math/tex; mode=display">\bex \left\{\ba{cc} -x^2-\cfrac{y^2}{4}+z^2=1\\ y=1 \ea\right. \ra \left\{\ba{cc} -\cfrac{x^2}{5/4}+\cfrac{z^2}{5/4}=1 \\ x=1 \ea\right. \mbox{ 是一条双曲线.} \eex</script>

8多元函数的微分学及其应用

8.1多元函数的基本概念

4(6) 求函数 \dps{z=\cfrac{\sqrt{x-y}}{\ln (1-x^2-y^2)}}<script type="math/tex">\dps{z=\cfrac{\sqrt{x-y}}{\ln (1-x^2-y^2)}}</script> 的定义域, 并画出其图形.

解答: 由 x-y\geq 0\ra x\geq y<script type="math/tex">x-y\geq 0\ra x\geq y</script> 及 \bex \ln (1-x^2-y^2)\neq 0 &\ra& 0<1-x^2-y^2\mbox{ 且 }1\neq 1-x^2-y^2\\ &\ra& 0<x^2+y^2<1 \eex<script type="math/tex; mode=display">\bex \ln (1-x^2-y^2)\neq 0 &\ra& 0<1-x^2-y^2\mbox{ 且 }1\neq 1-x^2-y^2\\ &\ra& 0\sed{(x,y)\in {\bf R}^2\ |\ x\geq y,\ 0<x^2+y^2<1}<script type="math/tex">\sed{(x,y)\in {\bf R}^2\ |\ x\geq y,\ 0

5(4) 求极限 \dps{\lim_{(x,y)\to (0,0)}\cfrac{2-\sqrt{xy+4}}{xy}}<script type="math/tex">\dps{\lim_{(x,y)\to (0,0)}\cfrac{2-\sqrt{xy+4}}{xy}}</script>.

解答: \bex \mbox{原极限} &=&\lim_{(x,y)\to (0,0)}\cfrac{2-\sqrt{xy-4}}{xy}\cdot\cfrac{2+\sqrt{xy+4}}{2+\sqrt{xy+4}}\\ &=&\lim_{(x,y)\to (0,0)}\cfrac{4-(xy+4)}{xy(2+\sqrt{xy+4})}\\ &=&-\lim_{(x,y)\to (0,0)}\cfrac{1}{2+\sqrt{xy+4}} =-\cfrac{1}{2+\sqrt{4}} =-\cfrac{1}{4}. \eex<script type="math/tex; mode=display">\bex \mbox{原极限} &=&\lim_{(x,y)\to (0,0)}\cfrac{2-\sqrt{xy-4}}{xy}\cdot\cfrac{2+\sqrt{xy+4}}{2+\sqrt{xy+4}}\\ &=&\lim_{(x,y)\to (0,0)}\cfrac{4-(xy+4)}{xy(2+\sqrt{xy+4})}\\ &=&-\lim_{(x,y)\to (0,0)}\cfrac{1}{2+\sqrt{xy+4}} =-\cfrac{1}{2+\sqrt{4}} =-\cfrac{1}{4}. \eex</script>

6(2) 证明极限 \dps{\lim_{(x,y)\to (0,0)}\cfrac{x^2}{x^2+y^2-x}}<script type="math/tex">\dps{\lim_{(x,y)\to (0,0)}\cfrac{x^2}{x^2+y^2-x}}</script> 不存在.

证明: 为证二元函数 \dps{f(x,y)=\cfrac{x^2}{x^2+y^2-x}}<script type="math/tex">\dps{f(x,y)=\cfrac{x^2}{x^2+y^2-x}}</script> 在 (0,0)<script type="math/tex">(0,0)</script> 处极限不存在, 仅须找到两条过 (0,0)<script type="math/tex">(0,0)</script> 的曲线 C_1<script type="math/tex">C_1</script>, C_2<script type="math/tex">C_2</script>, 使得 f(x,y)<script type="math/tex">f(x,y)</script> 沿着它们趋近于 (0,0)<script type="math/tex">(0,0)</script> 时极限不存在或者存在但不相等. 很容易想到 C_1:\ x=0<script type="math/tex">C_1:\ x=0</script>, 于其上 f(x,y)=0<script type="math/tex">f(x,y)=0</script>, 而 \dps{\lim_{(x,y)\to (0,0)\atop (x,y)\in C_1}f(x,y)=0}<script type="math/tex">\dps{\lim_{(x,y)\to (0,0)\atop (x,y)\in C_1}f(x,y)=0}</script>. 现再找一条曲线 C_2<script type="math/tex">C_2</script> 使 \dps{\lim_{(x,y)\to (0,0)\atop (x,y)\in C_2}f(x,y)=1}<script type="math/tex">\dps{\lim_{(x,y)\to (0,0)\atop (x,y)\in C_2}f(x,y)=1}</script>. 我们可以直接就让 \bex 1=f(x,y)=\cfrac{x^2}{x^2+y^2-x}\ra C_2:\ x=y^2. \eex<script type="math/tex; mode=display">\bex 1=f(x,y)=\cfrac{x^2}{x^2+y^2-x}\ra C_2:\ x=y^2. \eex</script> 综上所述, 我们有结论.

7(2) 试问函数 \dps{ f(x,y)=\left\{\ba{cc} \cfrac{\sin (xy)}{x},&x\neq 0\\ y,&x=0 \ea\right. }<script type="math/tex">\dps{ f(x,y)=\left\{\ba{cc} \cfrac{\sin (xy)}{x},&x\neq 0\\ y,&x=0 \ea\right. }</script> 是否在全平面 {\bf R}^2<script type="math/tex">{\bf R}^2</script> 内连续, 说明理由.

解答: 当 x\neq 0<script type="math/tex">x\neq 0</script> 时, \dps{f(x,y)=\cfrac{\sin(xy)}{x}}<script type="math/tex">\dps{f(x,y)=\cfrac{\sin(xy)}{x}}</script>, 这是初等函数, 而连续; 当 x=0<script type="math/tex">x=0</script> 时, 对任意的 y_0\in{\bf R}<script type="math/tex">y_0\in{\bf R}</script>, 有 \bex \lim_{(x,y)\to (0,y_0)}f(x,y) &=&\lim_{(x,y)\to (0,y_0)}\cfrac{\sin (xy)}{y} =\lim_{(x,y)\to (0,y_0)}\cfrac{\sin(xy)}{xy}\cdot y\\ &=&\lim_{t\to 0}\cfrac{\sin t}{t}\cdot\lim_{y\to y_0}y =1\cdot y_0=y_0=f(0,y_0). \eex<script type="math/tex; mode=display">\bex \lim_{(x,y)\to (0,y_0)}f(x,y) &=&\lim_{(x,y)\to (0,y_0)}\cfrac{\sin (xy)}{y} =\lim_{(x,y)\to (0,y_0)}\cfrac{\sin(xy)}{xy}\cdot y\\ &=&\lim_{t\to 0}\cfrac{\sin t}{t}\cdot\lim_{y\to y_0}y =1\cdot y_0=y_0=f(0,y_0). \eex</script> 而 f(x,y)<script type="math/tex">f(x,y)</script> 在全平面 {\bf R}^2<script type="math/tex">{\bf R}^2</script> 上连续.

8.2 偏导数

1(12). 求 u=\arctan(x+y)^z<script type="math/tex">u=\arctan(x+y)^z</script> 的偏导数.

解答: \bex u_x=\cfrac{1}{1+\sez{(x+y)^z}^2}\cdot z(x+y)^{z-1}\cdot 1 =\cfrac{z(x+y)^{z-1}}{1+(x+y)^{2z}}; \eex<script type="math/tex; mode=display">\bex u_x=\cfrac{1}{1+\sez{(x+y)^z}^2}\cdot z(x+y)^{z-1}\cdot 1 =\cfrac{z(x+y)^{z-1}}{1+(x+y)^{2z}}; \eex</script> \bex u_y=\cfrac{z(x+y)^{z-1}}{1+(x+y)^{2z}}\quad\sex{\mbox{由 } x,y\mbox{ 的对称性}}; \eex<script type="math/tex; mode=display">\bex u_y=\cfrac{z(x+y)^{z-1}}{1+(x+y)^{2z}}\quad\sex{\mbox{由 } x,y\mbox{ 的对称性}}; \eex</script> \bex u_z=\cfrac{1}{1+\sez{(x+y)^z}^2}\cdot (x+y)^z\ln (x+y) =\cfrac{(x+y)^z\ln (x+y)}{1+(x+y)^{2z}}. \eex<script type="math/tex; mode=display">\bex u_z=\cfrac{1}{1+\sez{(x+y)^z}^2}\cdot (x+y)^z\ln (x+y) =\cfrac{(x+y)^z\ln (x+y)}{1+(x+y)^{2z}}. \eex</script>

5. 设 \dps{f(x,y)=x+(y-1)\arcsin\sqrt{\cfrac{x}{y}}}<script type="math/tex">\dps{f(x,y)=x+(y-1)\arcsin\sqrt{\cfrac{x}{y}}}</script>, 求 f_x(x,1)<script type="math/tex">f_x(x,1)</script>.

解答: \bex f_x(x,y)=1+(y-1)\sez{\arcsin\sqrt{\cfrac{x}{y}}\ }_x; \eex<script type="math/tex; mode=display">\bex f_x(x,y)=1+(y-1)\sez{\arcsin\sqrt{\cfrac{x}{y}}\ }_x; \eex</script> \bex f_x(x,1)=1+0\cdot \left.\sez{\arcsin\sqrt{\cfrac{x}{y}}\ }_x\right|_{x=x\atop y=1} =1. \eex<script type="math/tex; mode=display">\bex f_x(x,1)=1+0\cdot \left.\sez{\arcsin\sqrt{\cfrac{x}{y}}\ }_x\right|_{x=x\atop y=1} =1. \eex</script>

7. 求曲线 \dps{\left\{\ba{cc} z=\cfrac{x^2+y^2}{4}\\ y=4 \ea\right.}<script type="math/tex">\dps{\left\{\ba{cc} z=\cfrac{x^2+y^2}{4}\\ y=4 \ea\right.}</script> 在点 (2,4,5)<script type="math/tex">(2,4,5)</script> 处的切线对于 x<script type="math/tex">x</script> 轴的倾角 \theta<script type="math/tex">\theta</script>.

解答: 注意到曲线 \dps{\left\{\ba{cc} z=\cfrac{x^2+y^2}{4}\\ y=4 \ea\right.}<script type="math/tex">\dps{\left\{\ba{cc} z=\cfrac{x^2+y^2}{4}\\ y=4 \ea\right.}</script> 就是二元函数 \dps{z=\cfrac{x^2+y^2}{4}}<script type="math/tex">\dps{z=\cfrac{x^2+y^2}{4}}</script> 与平面 y=4<script type="math/tex">y=4</script> 的交线, 其在 (2,4,5)<script type="math/tex">(2,4,5)</script> 处的切线的斜率就是 \dps{z=\cfrac{x^2+y^2}{4}}<script type="math/tex">\dps{z=\cfrac{x^2+y^2}{4}}</script> 关于 x<script type="math/tex">x</script> 的偏导数在 (2,4)<script type="math/tex">(2,4)</script> 处的值, 于是 \bex \tan\theta=k=z_x|_{x=2\atop y=4} =\left.\cfrac{x}{2}\right|_{x=2\atop y=4} =1, \eex<script type="math/tex; mode=display">\bex \tan\theta=k=z_x|_{x=2\atop y=4} =\left.\cfrac{x}{2}\right|_{x=2\atop y=4} =1, \eex</script> \bex \theta=\cfrac{\pi}{4}. \eex<script type="math/tex; mode=display">\bex \theta=\cfrac{\pi}{4}. \eex</script>

10. 设 z=xe^{xy}<script type="math/tex">z=xe^{xy}</script>, 求 z_{xyx}<script type="math/tex">z_{xyx}</script> 与 z_{xyy}<script type="math/tex">z_{xyy}</script>.

解答: \bex z_x=1\cdot e^{xy}+x\cdot e^{xy}y=(1+xy)e^{xy}, \eex<script type="math/tex; mode=display">\bex z_x=1\cdot e^{xy}+x\cdot e^{xy}y=(1+xy)e^{xy}, \eex</script> \bex z_{xy} =(z_x)_y=x\cdot e^{xy}+(1+xy)\cdot e^{xy}x =(2x+x^2y)e^{xy}, \eex<script type="math/tex; mode=display">\bex z_{xy} =(z_x)_y=x\cdot e^{xy}+(1+xy)\cdot e^{xy}x =(2x+x^2y)e^{xy}, \eex</script> \bex z_{xyx} =(z_{xy})_x =(2+2xy)\cdot e^{xy} +(2x+x^2y)\cdot e^{xy}y =(2+4xy+x^2y^2)e^{xy}, \eex<script type="math/tex; mode=display">\bex z_{xyx} =(z_{xy})_x =(2+2xy)\cdot e^{xy} +(2x+x^2y)\cdot e^{xy}y =(2+4xy+x^2y^2)e^{xy}, \eex</script> \bex z_{xyy} =(z_{xy})_y =x^2\cdot e^{xy}+(2x+x^2y)\cdot e^{xy}x =(3x^2+x^3y)e^{xy}. \eex<script type="math/tex; mode=display">\bex z_{xyy} =(z_{xy})_y =x^2\cdot e^{xy}+(2x+x^2y)\cdot e^{xy}x =(3x^2+x^3y)e^{xy}. \eex</script>

8.3 全微分

1(5). 求函数 z=x^{yz}<script type="math/tex">z=x^{yz}</script> 的全微分.

解答: 由 \bex u_x=yzx^{yz-1}, \eex<script type="math/tex; mode=display">\bex u_x=yzx^{yz-1}, \eex</script> \bex u_y=\sez{(x^z)^y}_y =(x^z)^y\cdot \ln x^z =x^{yz}z\ln x, \eex<script type="math/tex; mode=display">\bex u_y=\sez{(x^z)^y}_y =(x^z)^y\cdot \ln x^z =x^{yz}z\ln x, \eex</script> \bex u_z=x^{yz}y\ln x, \eex<script type="math/tex; mode=display">\bex u_z=x^{yz}y\ln x, \eex</script> 知 \bex du=u_x\rd x+u_y\rd y+u_z\rd z =x^{yz-1}\sex{yz\rd x+zx\ln x \rd y +xy\ln x \rd z}. \eex<script type="math/tex; mode=display">\bex du=u_x\rd x+u_y\rd y+u_z\rd z =x^{yz-1}\sex{yz\rd x+zx\ln x \rd y +xy\ln x \rd z}. \eex</script>

2. 求函数 \dps{z=y/x}<script type="math/tex">\dps{z=y/x}</script> 当 x=2, y=1, \lap x=0.1, \lap y=-0.2<script type="math/tex">x=2, y=1, \lap x=0.1, \lap y=-0.2</script> 时的全增量与全微分.

解答: 全增量 \bex \lap z=z(x+\lap x,y+\lap y)-z(x,y) =\cfrac{1-0.2}{2+0.1}-\cfrac{1}{2} =-\cfrac{5}{42}. \eex<script type="math/tex; mode=display">\bex \lap z=z(x+\lap x,y+\lap y)-z(x,y) =\cfrac{1-0.2}{2+0.1}-\cfrac{1}{2} =-\cfrac{5}{42}. \eex</script> 全微分 \bex \rd z|_{x=2,y=1\atop \lap x=0.1,\lap y=-0.2} &=&z_x|_{x=2\atop y=1}\rd x|_{\lap x=0.1} +z_y|_{x=2\atop y=1}\rd y|_{\lap y=0-0.2}\\ &=&\left.-\cfrac{y}{x^2}\right|_{x=2\atop y=1} \cdot 0.1 +\left.\cfrac{1}{x}\right|_{x=2\atop y=1} \cdot (-0.2)\\ &=&-\cfrac{1}{4}\cdot 0.1+\cfrac{1}{2}\cdot (-0.2) =-0.125. \eex<script type="math/tex; mode=display">\bex \rd z|_{x=2,y=1\atop \lap x=0.1,\lap y=-0.2} &=&z_x|_{x=2\atop y=1}\rd x|_{\lap x=0.1} +z_y|_{x=2\atop y=1}\rd y|_{\lap y=0-0.2}\\ &=&\left.-\cfrac{y}{x^2}\right|_{x=2\atop y=1} \cdot 0.1 +\left.\cfrac{1}{x}\right|_{x=2\atop y=1} \cdot (-0.2)\\ &=&-\cfrac{1}{4}\cdot 0.1+\cfrac{1}{2}\cdot (-0.2) =-0.125. \eex</script>

4(3). 求函数 \dps{u=e^x\sin (yz)}<script type="math/tex">\dps{u=e^x\sin (yz)}</script> 在点 \dps{\sex{1,\cfrac{1}{2},\cfrac{\pi}{2}}}<script type="math/tex">\dps{\sex{1,\cfrac{1}{2},\cfrac{\pi}{2}}}</script>　处的全微分．

解答: 由 \bex u_x=e^x\sin (yz),\quad u_y=ze^x\cos(yz),\quad u_z=ye^x\cos(yz) \eex<script type="math/tex; mode=display">\bex u_x=e^x\sin (yz),\quad u_y=ze^x\cos(yz),\quad u_z=ye^x\cos(yz) \eex</script> 知 \bex \rd z|_{x=1, y=1/2, z=\pi/2} =\cfrac{\sqrt{2}}{2} \sex{\rd x+\cfrac{\pi}{2}\rd y+\cfrac{1}{2}\rd z}. \eex<script type="math/tex; mode=display">\bex \rd z|_{x=1, y=1/2, z=\pi/2} =\cfrac{\sqrt{2}}{2} \sex{\rd x+\cfrac{\pi}{2}\rd y+\cfrac{1}{2}\rd z}. \eex</script>

8.4多元复合函数的求导法则

4. 设 z=e^{x-2y}<script type="math/tex">z=e^{x-2y}</script>, 而 x=\sin t,\ y=t^2<script type="math/tex">x=\sin t,\ y=t^2</script>, 求 \dps{\cfrac{\rd z}{dt}}<script type="math/tex">\dps{\cfrac{\rd z}{dt}}</script>.

解答: 变量之间的关系图为 \bex z\left<\ba{cc} x\cdots\cdots t\\ y\cdots\cdots t\ea\right. \eex<script type="math/tex; mode=display">\bex z\left<\ba{cc} x\cdots\cdots t\\ y\cdots\cdots t\ea\right. \eex</script> 而 \bex z_t&=&z_x\cdot x_t+z_y\cdot y_t\\ &=&e^{x-2y}\cdot \cos t+e^{x-2y}(-2)\cdot 3t^2\\ &=&e^{\sin t-2t^3}(\cos t-6t^2). \eex<script type="math/tex; mode=display">\bex z_t&=&z_x\cdot x_t+z_y\cdot y_t\\ &=&e^{x-2y}\cdot \cos t+e^{x-2y}(-2)\cdot 3t^2\\ &=&e^{\sin t-2t^3}(\cos t-6t^2). \eex</script>

10. 设 \dps{z=\arctan \cfrac{x}{y}}<script type="math/tex">\dps{z=\arctan \cfrac{x}{y}}</script>, 而 x=u+v,\ y=u-v<script type="math/tex">x=u+v,\ y=u-v</script>, 求证: \dps{z_u+z_v=\cfrac{u-v}{u^2+v^2}}<script type="math/tex">\dps{z_u+z_v=\cfrac{u-v}{u^2+v^2}}</script>.

证明: 变量之间的关系图为 \bex z\left<\ba{cc} x\left<\ba{cc}u\\v\ea\right.\\ y\left<\ba{cc}u\\v\ea\right. \ea\right. \eex<script type="math/tex; mode=display">\bex z\left<\ba{cc} x\left<\ba{cc}u\\v\ea\right.\\ y\left<\ba{cc}u\\v\ea\right. \ea\right. \eex</script> 而 \bex z_u&=&z_x\cdot x_u+z_y\cdot y_u\\ &=&\cfrac{1}{1+\sex{\cfrac{x}{y}}^2}\cfrac{1}{y}\cdot 1 +\cfrac{1}{1+\sex{\cfrac{x}{y}}^2}\sex{-\cfrac{x}{y^2}}\cdot 1\\ &=&\cfrac{y-x}{x^2+y^2}=\cfrac{-v}{u^2+v^2}, \eex<script type="math/tex; mode=display">\bex z_u&=&z_x\cdot x_u+z_y\cdot y_u\\ &=&\cfrac{1}{1+\sex{\cfrac{x}{y}}^2}\cfrac{1}{y}\cdot 1 +\cfrac{1}{1+\sex{\cfrac{x}{y}}^2}\sex{-\cfrac{x}{y^2}}\cdot 1\\ &=&\cfrac{y-x}{x^2+y^2}=\cfrac{-v}{u^2+v^2}, \eex</script> \bex z_v&=&z_x\cdot x_v+z_y\cdot y_v\\ &=&\cfrac{1}{1+\sex{\cfrac{x}{y}}^2}\cfrac{1}{y}\cdot 1 +\cfrac{1}{1+\sex{\cfrac{x}{y}}^2}\sex{-\cfrac{x}{y^2}}\cdot (-1)\\ &=&\cfrac{y+x}{x^2+y^2} =\cfrac{u}{u^2+v^2}, \eex<script type="math/tex; mode=display">\bex z_v&=&z_x\cdot x_v+z_y\cdot y_v\\ &=&\cfrac{1}{1+\sex{\cfrac{x}{y}}^2}\cfrac{1}{y}\cdot 1 +\cfrac{1}{1+\sex{\cfrac{x}{y}}^2}\sex{-\cfrac{x}{y^2}}\cdot (-1)\\ &=&\cfrac{y+x}{x^2+y^2} =\cfrac{u}{u^2+v^2}, \eex</script> \bex z_u+z_v=\cfrac{u-v}{u^2+v^2}. \eex<script type="math/tex; mode=display">\bex z_u+z_v=\cfrac{u-v}{u^2+v^2}. \eex</script>

12. 设 f(u)<script type="math/tex">f(u)</script> 具有连续导数, \dps{z=xy+xf\sex{\cfrac{y}{x}}}<script type="math/tex">\dps{z=xy+xf\sex{\cfrac{y}{x}}}</script>, 求证 \dps{xz_x+yz_y=xy+z}<script type="math/tex">\dps{xz_x+yz_y=xy+z}</script>.

解答: 变量之间的关系图为 \bex z\left<\ba{lll} x\\ y\\ u\left<\ba{ll}x\\y\ea\right. \ea\right. \quad z=xy+xf(u),\ u=\cfrac{y}{x}. \eex<script type="math/tex; mode=display">\bex z\left<\ba{lll} x\\ y\\ u\left<\ba{ll}x\\y\ea\right. \ea\right. \quad z=xy+xf(u),\ u=\cfrac{y}{x}. \eex</script> 于是 \bex z_x=\sez{y+f(u)}+xf‘(u)\cdot\sex{-\cfrac{y}{x^2}} =y+f\sex{\cfrac{y}{x}}-\cfrac{y}{x^3}f‘\sex{\cfrac{y}{x}}, \eex<script type="math/tex; mode=display">\bex z_x=\sez{y+f(u)}+xf‘(u)\cdot\sex{-\cfrac{y}{x^2}} =y+f\sex{\cfrac{y}{x}}-\cfrac{y}{x^3}f‘\sex{\cfrac{y}{x}}, \eex</script> \bex z_y=x+xf‘(u)\cdot \cfrac{1}{x} =x+f‘\sex{\cfrac{y}{x}}, \eex<script type="math/tex; mode=display">\bex z_y=x+xf‘(u)\cdot \cfrac{1}{x} =x+f‘\sex{\cfrac{y}{x}}, \eex</script> \bex xz_x+yz_y &=&\sez{xy+xf\sex{\cfrac{y}{x}}-\cfrac{y}{x^2}f‘\sex{\cfrac{y}{x}}} +\sez{xy+yf‘\sex{\cfrac{y}{x}}}\\ &=&xy+z. \eex<script type="math/tex; mode=display">\bex xz_x+yz_y &=&\sez{xy+xf\sex{\cfrac{y}{x}}-\cfrac{y}{x^2}f‘\sex{\cfrac{y}{x}}} +\sez{xy+yf‘\sex{\cfrac{y}{x}}}\\ &=&xy+z. \eex</script>

14(2). 设 f<script type="math/tex">f</script> 具有二阶连续偏导数, 求函数 \dps{z=f\sex{xy,\cfrac{x}{y}}}<script type="math/tex">\dps{z=f\sex{xy,\cfrac{x}{y}}}</script> 的 \dps{z_{xx},z_{xy},z_{yy}}<script type="math/tex">\dps{z_{xx},z_{xy},z_{yy}}</script>.

解答: 变量之间的关系图为 \bex z\left<\ba{cc} u\left<\ba{cc}x\\y\ea\right.\\ v\left<\ba{cc}x\\y\ea\right. \ea\right. \quad z=f(u,v),\ u=xy,\ v=\cfrac{x}{y}. \eex<script type="math/tex; mode=display">\bex z\left<\ba{cc} u\left<\ba{cc}x\\y\ea\right.\\ v\left<\ba{cc}x\\y\ea\right. \ea\right. \quad z=f(u,v),\ u=xy,\ v=\cfrac{x}{y}. \eex</script> 由 \bex \left\{ \ba{ll} u_x=y\\ u_y=x \ea\right. \quad \left\{ \ba{ll} v_x=\cfrac{1}{y}\\ v_y=-\cfrac{x}{y^2} \ea\right. \eex<script type="math/tex; mode=display">\bex \left\{ \ba{ll} u_x=y\\ u_y=x \ea\right. \quad \left\{ \ba{ll} v_x=\cfrac{1}{y}\\ v_y=-\cfrac{x}{y^2} \ea\right. \eex</script> 知 \bex z_x=f_1‘\cdot y+f_2‘\cdot\cfrac{1}{y}, \eex<script type="math/tex; mode=display">\bex z_x=f_1‘\cdot y+f_2‘\cdot\cfrac{1}{y}, \eex</script> \bex z_y=f_1‘\cdot x+f_2‘\cdot \sex{-\cfrac{x}{y^2}}; \eex<script type="math/tex; mode=display">\bex z_y=f_1‘\cdot x+f_2‘\cdot \sex{-\cfrac{x}{y^2}}; \eex</script> \bex z_{xx} &=&\sez{f_{11}‘‘\cdot y+f_{12}‘‘\cdot \cfrac{1}{y}}\cdot y+\sez{f_{21}‘‘\cdot y+f_{22}‘‘\cdot\cfrac{1}{y}}\cdot\cfrac{1}{y}\\ &=&y^2f_{11}‘‘+2f_{12}‘‘+\cfrac{1}{y^2}f_{22}‘‘, \eex<script type="math/tex; mode=display">\bex z_{xx} &=&\sez{f_{11}‘‘\cdot y+f_{12}‘‘\cdot \cfrac{1}{y}}\cdot y+\sez{f_{21}‘‘\cdot y+f_{22}‘‘\cdot\cfrac{1}{y}}\cdot\cfrac{1}{y}\\ &=&y^2f_{11}‘‘+2f_{12}‘‘+\cfrac{1}{y^2}f_{22}‘‘, \eex</script> \bex z_{xy} &=&\sez{f_{11}‘‘\cdot x+f_{12}‘‘\cdot\sex{-\cfrac{x}{y^2}}}\cdot y+f_1‘\cdot 1\\ & & +\sez{f_{21}‘‘\cdot x+f_{22}‘‘\cdot\sex{-\cfrac{x}{y^2}}}\cdot\cfrac{1}{y}+f_2‘\cdot \cfrac{-1}{y^2}\\ &=&f_1‘-\cfrac{1}{y^2}f_2‘ +xyf_{11}‘ -\cfrac{x}{y^3}f_{22}‘‘, \eex<script type="math/tex; mode=display">\bex z_{xy} &=&\sez{f_{11}‘‘\cdot x+f_{12}‘‘\cdot\sex{-\cfrac{x}{y^2}}}\cdot y+f_1‘\cdot 1\\ & & +\sez{f_{21}‘‘\cdot x+f_{22}‘‘\cdot\sex{-\cfrac{x}{y^2}}}\cdot\cfrac{1}{y}+f_2‘\cdot \cfrac{-1}{y^2}\\ &=&f_1‘-\cfrac{1}{y^2}f_2‘ +xyf_{11}‘ -\cfrac{x}{y^3}f_{22}‘‘, \eex</script> \bex z_{yy} &=&\sez{f_{11}‘‘\cdot x+f_{12}‘‘\cdot\sex{-\cfrac{x}{y^2}}}\cdot x\\ & & +\sez{f_{21}‘‘\cdot x+f_{22}‘‘\cdot\sex{-\cfrac{x}{y^2}}}\cdot \sex{-\cfrac{x}{y^2}} +f_2‘\cdot \cfrac{2x}{y^3}\\ &=&\cfrac{2x}{y^3}f_2‘ +x^2f_{11}‘‘ -\cfrac{2x^2}{y^2}f_{12}‘‘ +\cfrac{x^2}{y^4}f_{22}‘‘. \eex<script type="math/tex; mode=display">\bex z_{yy} &=&\sez{f_{11}‘‘\cdot x+f_{12}‘‘\cdot\sex{-\cfrac{x}{y^2}}}\cdot x\\ & & +\sez{f_{21}‘‘\cdot x+f_{22}‘‘\cdot\sex{-\cfrac{x}{y^2}}}\cdot \sex{-\cfrac{x}{y^2}} +f_2‘\cdot \cfrac{2x}{y^3}\\ &=&\cfrac{2x}{y^3}f_2‘ +x^2f_{11}‘‘ -\cfrac{2x^2}{y^2}f_{12}‘‘ +\cfrac{x^2}{y^4}f_{22}‘‘. \eex</script>

8.5 隐函数的求导公式

1(3). 设 x^y=y^x<script type="math/tex">x^y=y^x</script> 确定 y<script type="math/tex">y</script> 是 x<script type="math/tex">x</script> 的函数, 求 \dps{\cfrac{\rd y}{\rd x}}<script type="math/tex">\dps{\cfrac{\rd y}{\rd x}}</script>.

解答: 由方程 x^y=y^x<script type="math/tex">x^y=y^x</script> 知 \bex y\ln x=x\ln y. \eex<script type="math/tex; mode=display">\bex y\ln x=x\ln y. \eex</script> 两边对 x<script type="math/tex">x</script> 求导有 \bex y‘\ln x+y\cdot \cfrac{1}{x} =\ln y+x\cdot \cfrac{1}{y}y‘, \eex<script type="math/tex; mode=display">\bex y‘\ln x+y\cdot \cfrac{1}{x} =\ln y+x\cdot \cfrac{1}{y}y‘, \eex</script> \bex y‘=\cfrac{\ln y-\cfrac{y}{x}}{\ln x-\cfrac{x}{y}} =\cfrac{y\sex{x\ln y-y}}{x\sex{y\ln x-x}}. \eex<script type="math/tex; mode=display">\bex y‘=\cfrac{\ln y-\cfrac{y}{x}}{\ln x-\cfrac{x}{y}} =\cfrac{y\sex{x\ln y-y}}{x\sex{y\ln x-x}}. \eex</script>

2(4). 设 \sin z=xyz<script type="math/tex">\sin z=xyz</script> 确定 z<script type="math/tex">z</script> 是 x,y<script type="math/tex">x,y</script> 的函数, 求 \dps{z_x,\ z_y}<script type="math/tex">\dps{z_x,\ z_y}</script>.

解答: 对 \sin z=xyz<script type="math/tex">\sin z=xyz</script> 两边分别关于 x,y<script type="math/tex">x,y</script> 求偏导有 \bex \cos z\cdot z_x=yz+xy\cdot z_x,\quad \cos z\cdot z_y=xz+xy\cdot z_y, \eex<script type="math/tex; mode=display">\bex \cos z\cdot z_x=yz+xy\cdot z_x,\quad \cos z\cdot z_y=xz+xy\cdot z_y, \eex</script> 于是 \bex z_x=\cfrac{yz}{\cos z-xy},\quad z_y=\cfrac{xz}{\cos z-xy}. \eex<script type="math/tex; mode=display">\bex z_x=\cfrac{yz}{\cos z-xy},\quad z_y=\cfrac{xz}{\cos z-xy}. \eex</script> {5.} 设函数 \varphi(u,v)<script type="math/tex">\varphi(u,v)</script> 具有连续偏导数, 且方程 \varphi(cx-az,cy-bz)=0<script type="math/tex">\varphi(cx-az,cy-bz)=0</script> 确定了函数 z=z(x,y)<script type="math/tex">z=z(x,y)</script>. 求证: \dps{az_x+bz_y=c}<script type="math/tex">\dps{az_x+bz_y=c}</script>.

证明: 对 \varphi(cx-az,cy-bz)=0<script type="math/tex">\varphi(cx-az,cy-bz)=0</script> 两边分别关于 x,y<script type="math/tex">x,y</script> 求偏导有 \bex \varphi_u\cdot (c-az_x)+\varphi_v\cdot(-bz_x)=0,\quad \varphi_u\cdot(-az_y)+\varphi_v\cdot(c-bz_y)=0, \eex<script type="math/tex; mode=display">\bex \varphi_u\cdot (c-az_x)+\varphi_v\cdot(-bz_x)=0,\quad \varphi_u\cdot(-az_y)+\varphi_v\cdot(c-bz_y)=0, \eex</script> \bex z_x=\cfrac{c\varphi_u}{a\varphi_u+b\varphi_v},\quad z_y=\cfrac{c\varphi_v}{a\varphi_u+b\varphi_v}, \eex<script type="math/tex; mode=display">\bex z_x=\cfrac{c\varphi_u}{a\varphi_u+b\varphi_v},\quad z_y=\cfrac{c\varphi_v}{a\varphi_u+b\varphi_v}, \eex</script> \bex az_x+bz_y=\cfrac{c\sex{a\varphi_u+b\varphi_v}}{a\varphi_u+b\varphi_v}=c. \eex<script type="math/tex; mode=display">\bex az_x+bz_y=\cfrac{c\sex{a\varphi_u+b\varphi_v}}{a\varphi_u+b\varphi_v}=c. \eex</script> {8(3).} 求由方程组 \dps{\left\{\ba{ll} u^3+xv-y=0\\ v^3+yu-x=0 \ea\right.}<script type="math/tex">\dps{\left\{\ba{ll} u^3+xv-y=0\\ v^3+yu-x=0 \ea\right.}</script> 所确定的函数 u,v<script type="math/tex">u,v</script> 的偏导数 \dps{\cfrac{\p u}{\p x},\ \cfrac{\p v}{\p x}}<script type="math/tex">\dps{\cfrac{\p u}{\p x},\ \cfrac{\p v}{\p x}}</script>.

解答: 令 F(x,y,u,v)=u^3+xv-y,\ G(x,y,u,v)=v^3+yu-v<script type="math/tex">F(x,y,u,v)=u^3+xv-y,\ G(x,y,u,v)=v^3+yu-v</script>. 则 \bex \ba{cccc} F_x=v,&F_y=-1,&F_u=3u^2,&F_v=x;\\ G_x=-1,&G_y=u,&G_u=y,&G_v=3v^2. \ea \eex<script type="math/tex; mode=display">\bex \ba{cccc} F_x=v,&F_y=-1,&F_u=3u^2,&F_v=x;\\ G_x=-1,&G_y=u,&G_u=y,&G_v=3v^2. \ea \eex</script> Jacobi<script type="math/tex">Jacobi</script> 式为 \bex J=\cfrac{\p (F,G)}{\p (u,v)} =\sev{\ba{cc} 3u^2&x\\ y&3v^2 \ea} =9u^2v^2-xy. \eex<script type="math/tex; mode=display">\bex J=\cfrac{\p (F,G)}{\p (u,v)} =\sev{\ba{cc} 3u^2&x\\ y&3v^2 \ea} =9u^2v^2-xy. \eex</script> 于是 \bex \cfrac{\p u}{\p x} =-\cfrac{1}{J}\cfrac{\p (F,G)}{\p x,v} =-\cfrac{1}{9u^2v^2-xy}\sev{\ba{cc} v&x\\ -1&3v^2 \ea} =-\cfrac{3v^3+x}{9u^2v^2-xy}; \eex<script type="math/tex; mode=display">\bex \cfrac{\p u}{\p x} =-\cfrac{1}{J}\cfrac{\p (F,G)}{\p x,v} =-\cfrac{1}{9u^2v^2-xy}\sev{\ba{cc} v&x\\ -1&3v^2 \ea} =-\cfrac{3v^3+x}{9u^2v^2-xy}; \eex</script> \bex \cfrac{\p v}{\p x} =-\cfrac{1}{J}\cfrac{\p (F,G)}{\p u,x} =-\cfrac{1}{9u^2v^2-xy}\sev{\ba{cc} 3u^2&v\\ y&-1 \ea} =\cfrac{3u^2+yv}{9u^2v^2-xy}. \eex<script type="math/tex; mode=display">\bex \cfrac{\p v}{\p x} =-\cfrac{1}{J}\cfrac{\p (F,G)}{\p u,x} =-\cfrac{1}{9u^2v^2-xy}\sev{\ba{cc} 3u^2&v\\ y&-1 \ea} =\cfrac{3u^2+yv}{9u^2v^2-xy}. \eex</script>

8.6 多元函数微分学的几何应用

1(4). 求曲线 \dps{x=\cfrac{2t}{1+t},\ y=\cfrac{1-t}{t},\ z=\sqrt{t}}<script type="math/tex">\dps{x=\cfrac{2t}{1+t},\ y=\cfrac{1-t}{t},\ z=\sqrt{t}}</script> 在点 (1,0,1)<script type="math/tex">(1,0,1)</script> 处的切线与法平面.

解答: 由 \sqrt{t}=z=1<script type="math/tex">\sqrt{t}=z=1</script> 知 t=1<script type="math/tex">t=1</script>. 又由 \bex x‘(t)=\cfrac{2}{(1+t)^2},\quad y‘(t)=-\cfrac{1}{t^2},\quad z‘(t)=\cfrac{1}{2\sqrt{t}} \eex<script type="math/tex; mode=display">\bex x‘(t)=\cfrac{2}{(1+t)^2},\quad y‘(t)=-\cfrac{1}{t^2},\quad z‘(t)=\cfrac{1}{2\sqrt{t}} \eex</script> 知曲线在点 (1,0,1)<script type="math/tex">(1,0,1)</script> 处的切向量为 \bex \sed{x‘(t),y‘(t),z‘(t)}|_{t=1} =\sed{\cfrac{1}{2},-1,\cfrac{1}{2}}, \eex<script type="math/tex; mode=display">\bex \sed{x‘(t),y‘(t),z‘(t)}|_{t=1} =\sed{\cfrac{1}{2},-1,\cfrac{1}{2}}, \eex</script> 或者 \bex \sed{1,-2,1}. \eex<script type="math/tex; mode=display">\bex \sed{1,-2,1}. \eex</script> 于是切线方程为 \bex \cfrac{x-1}{1} =\cfrac{y}{-2} =\cfrac{z-1}{1}; \eex<script type="math/tex; mode=display">\bex \cfrac{x-1}{1} =\cfrac{y}{-2} =\cfrac{z-1}{1}; \eex</script> 法平面方程为 \bex (x-1)-2y+(z-1)=0, \eex<script type="math/tex; mode=display">\bex (x-1)-2y+(z-1)=0, \eex</script> 或者 \bex x-2y+z-2=0. \eex<script type="math/tex; mode=display">\bex x-2y+z-2=0. \eex</script>

3(1). 求曲面 \dps{z=y+\ln \cfrac{x}{z}}<script type="math/tex">\dps{z=y+\ln \cfrac{x}{z}}</script> 在点 (1,1,1)<script type="math/tex">(1,1,1)</script> 处的切平面与法线.

解答: 记 \dps{F(x,y,z)=y+\ln \cfrac{x}{z}-z=y+\ln x-\ln z-z}<script type="math/tex">\dps{F(x,y,z)=y+\ln \cfrac{x}{z}-z=y+\ln x-\ln z-z}</script>, 则由 \bex F_x=\cfrac{1}{x},\quad F_y=1,\quad F_z=-\cfrac{1}{z}-1 \eex<script type="math/tex; mode=display">\bex F_x=\cfrac{1}{x},\quad F_y=1,\quad F_z=-\cfrac{1}{z}-1 \eex</script> 知曲面在 (1,1,1)<script type="math/tex">(1,1,1)</script> 处的法向量为 \bex {\bf n}=\sed{1,1,-2}. \eex<script type="math/tex; mode=display">\bex {\bf n}=\sed{1,1,-2}. \eex</script> 于是切平面方程为 \bex 1\cdot (x-1)+1\cdot (y-1)-2\cdot (z-1)=0, \eex<script type="math/tex; mode=display">\bex 1\cdot (x-1)+1\cdot (y-1)-2\cdot (z-1)=0, \eex</script> 或者 \bex x+y-2z=0; \eex<script type="math/tex; mode=display">\bex x+y-2z=0; \eex</script> 法线方程为 \bex \cfrac{x-1}{1} =\cfrac{y-1}{1} =\cfrac{z-1}{-2}. \eex<script type="math/tex; mode=display">\bex \cfrac{x-1}{1} =\cfrac{y-1}{1} =\cfrac{z-1}{-2}. \eex</script>

6. 求旋转椭球面 3x^2+y^2+z^2=16<script type="math/tex">3x^2+y^2+z^2=16</script> 上一点 (-1,-2,3)<script type="math/tex">(-1,-2,3)</script> 处的切平面与 xOy<script type="math/tex">xOy</script> 面对夹角的余弦.

解答: 记 F(x,y,z)=3x^2+y^2+z^2-16<script type="math/tex">F(x,y,z)=3x^2+y^2+z^2-16</script>, 则由 \bex F_x=6x,\quad F_y=2y,\quad F_z=2z \eex<script type="math/tex; mode=display">\bex F_x=6x,\quad F_y=2y,\quad F_z=2z \eex</script> 知曲面在 (-1,-2,3)<script type="math/tex">(-1,-2,3)</script> 处的法向量为 \bex {\bf n}_1=\sed{-6,-4,6}/2=\sed{-3,-2,3}, \eex<script type="math/tex; mode=display">\bex {\bf n}_1=\sed{-6,-4,6}/2=\sed{-3,-2,3}, \eex</script> 而与 xOy<script type="math/tex">xOy</script> 面的法向量为 \bex {\bf n}_2=\sed{0,0,1}. \eex<script type="math/tex; mode=display">\bex {\bf n}_2=\sed{0,0,1}. \eex</script> 于是两平面夹角的余弦为 \bex \cfrac{\sev{{\bf n}_1\cdot{\bf n}_2}}{\sev{{\bf n}_1}\cdot\sev{{\bf n}_2}} =\cfrac{3}{\sqrt{22}}. \eex<script type="math/tex; mode=display">\bex \cfrac{\sev{{\bf n}_1\cdot{\bf n}_2}}{\sev{{\bf n}_1}\cdot\sev{{\bf n}_2}} =\cfrac{3}{\sqrt{22}}. \eex</script>

7. 试证明曲面 \sqrt{x}+\sqrt{y}+\sqrt{z}=\sqrt{a}<script type="math/tex">\sqrt{x}+\sqrt{y}+\sqrt{z}=\sqrt{a}</script> 上任一点处的切平面在各坐标轴上的截距之和等于 a<script type="math/tex">a</script>.

证明: 设 F(x,y,z)=\sqrt{x}+\sqrt{y}+\sqrt{z}-\sqrt{a}<script type="math/tex">F(x,y,z)=\sqrt{x}+\sqrt{y}+\sqrt{z}-\sqrt{a}</script>, 则 \bex F_x=\cfrac{1}{2\sqrt{x}},\quad F_y=\cfrac{1}{2\sqrt{y}},\quad F_z=\cfrac{1}{2\sqrt{z}}. \eex<script type="math/tex; mode=display">\bex F_x=\cfrac{1}{2\sqrt{x}},\quad F_y=\cfrac{1}{2\sqrt{y}},\quad F_z=\cfrac{1}{2\sqrt{z}}. \eex</script> 于是曲面在点 (x_0,y_0,z_0)<script type="math/tex">(x_0,y_0,z_0)</script>　处的切平面方程为 \bex \cfrac{1}{2\sqrt{x_0}}(x-x_0) +\cfrac{1}{2\sqrt{y_0}}(y-y_0) +\cfrac{1}{2\sqrt{z_0}}(z-z_0) =0, \eex<script type="math/tex; mode=display">\bex \cfrac{1}{2\sqrt{x_0}}(x-x_0) +\cfrac{1}{2\sqrt{y_0}}(y-y_0) +\cfrac{1}{2\sqrt{z_0}}(z-z_0) =0, \eex</script> 或者 \bex \cfrac{x}{\sqrt{ax_0}} +\cfrac{y}{\sqrt{ay_0}} +\cfrac{z}{\sqrt{az_0}} =1. \eex<script type="math/tex; mode=display">\bex \cfrac{x}{\sqrt{ax_0}} +\cfrac{y}{\sqrt{ay_0}} +\cfrac{z}{\sqrt{az_0}} =1. \eex</script> 该平面在各坐标轴上的截距分别为 \bex \sqrt{ax_0},\quad \sqrt{ay_0},\quad \sqrt{az_0}, \eex<script type="math/tex; mode=display">\bex \sqrt{ax_0},\quad \sqrt{ay_0},\quad \sqrt{az_0}, \eex</script> 它们的和为 \bex \sqrt{a}\sex{\sqrt{x}+\sqrt{y}+\sqrt{z}} =\sqrt{a}\cdot\sqrt{a}=a. \eex<script type="math/tex; mode=display">\bex \sqrt{a}\sex{\sqrt{x}+\sqrt{y}+\sqrt{z}} =\sqrt{a}\cdot\sqrt{a}=a. \eex</script>

8.7 方向导数与梯度

2. 求函数 \dps{z=\cos(x+y)}<script type="math/tex">\dps{z=\cos(x+y)}</script> 在点 \dps{\sex{0,\cfrac{\pi}{2}}}<script type="math/tex">\dps{\sex{0,\cfrac{\pi}{2}}}</script> 处沿向量 {\bf l}=\sed{3,-4}<script type="math/tex">{\bf l}=\sed{3,-4}</script> 的方向的方向导数.

解答: 由 \bex z_x|_{\sex{0,\cfrac{\pi}{2}}} =-\sin(x+y)|_{\sex{0,\cfrac{\pi}{2}}} =-1, \eex<script type="math/tex; mode=display">\bex z_x|_{\sex{0,\cfrac{\pi}{2}}} =-\sin(x+y)|_{\sex{0,\cfrac{\pi}{2}}} =-1, \eex</script> \bex z_y|_{\sex{0,\cfrac{\pi}{2}}} =-\sin(x+y)|_{\sex{0,\cfrac{\pi}{2}}} =-1; \eex<script type="math/tex; mode=display">\bex z_y|_{\sex{0,\cfrac{\pi}{2}}} =-\sin(x+y)|_{\sex{0,\cfrac{\pi}{2}}} =-1; \eex</script> 及 \bex \cfrac{{\bf l}}{\sev{{\bf l}}} =\cfrac{\sed{3,-4}}{\sqrt{3^2+(-4)^2}} =\sed{\cfrac{3}{5},-\cfrac{4}{5}} \eex<script type="math/tex; mode=display">\bex \cfrac{{\bf l}}{\sev{{\bf l}}} =\cfrac{\sed{3,-4}}{\sqrt{3^2+(-4)^2}} =\sed{\cfrac{3}{5},-\cfrac{4}{5}} \eex</script> 知 \bex \cfrac{\p z}{\p {\bf l}} =-1\cdot\cfrac{3}{5} -1\cdot \sex{-\cfrac{4}{5}} =\cfrac{1}{5}. \eex<script type="math/tex; mode=display">\bex \cfrac{\p z}{\p {\bf l}} =-1\cdot\cfrac{3}{5} -1\cdot \sex{-\cfrac{4}{5}} =\cfrac{1}{5}. \eex</script>

10. 设 f(x,y,z)=x^2+2y^2+3z^2+xy+3x-26-6z<script type="math/tex">f(x,y,z)=x^2+2y^2+3z^2+xy+3x-26-6z</script>, 求 \grad f(0,0,0)<script type="math/tex">\grad f(0,0,0)</script> 及 \grad f(1,1,1)<script type="math/tex">\grad f(1,1,1)</script>.

解答: 由 \bex f_x=2x+y+3,\quad f_y=4y+x-2,\quad f_z=6z-6 \eex<script type="math/tex; mode=display">\bex f_x=2x+y+3,\quad f_y=4y+x-2,\quad f_z=6z-6 \eex</script> 知 \bex \grad f=\sed{2x+y+3,4y+x-2,6z-6}. \eex<script type="math/tex; mode=display">\bex \grad f=\sed{2x+y+3,4y+x-2,6z-6}. \eex</script> 于是 \bex \grad f(0,0,0)=\sed{3,-2,-6},\quad \grad f(1,1,1)=\sed{6,3,0}. \eex<script type="math/tex; mode=display">\bex \grad f(0,0,0)=\sed{3,-2,-6},\quad \grad f(1,1,1)=\sed{6,3,0}. \eex</script>

8.8 多元函数的极值问题

1. 求函数 f(x,y)=4(x-y)-x^2-y^2<script type="math/tex">f(x,y)=4(x-y)-x^2-y^2</script> 的极值.

解答: 由 \bex \left\{\ba{lll} f_x=4-2x=0&\ra&x=2\\ f_y=-4-2y=0&\ra&y=-2 \ea\right. \eex<script type="math/tex; mode=display">\bex \left\{\ba{lll} f_x=4-2x=0&\ra&x=2\\ f_y=-4-2y=0&\ra&y=-2 \ea\right. \eex</script> 知 (2,-2)<script type="math/tex">(2,-2)</script> 为 f<script type="math/tex">f</script> 的驻点. 又 \bex f_{xx}=-2,\quad f_{xy}=0,\quad f_{yy}=-2, \eex<script type="math/tex; mode=display">\bex f_{xx}=-2,\quad f_{xy}=0,\quad f_{yy}=-2, \eex</script> 我们有 \bex \sev{\ba{cc} f_{xx}&f_{xy}\\ f_{yx}&f_{yy} \ea} =4>0,\quad f_{xx}=-2<0. \eex<script type="math/tex; mode=display">\bex \sev{\ba{cc} f_{xx}&f_{xy}\\ f_{yx}&f_{yy} \ea} =4>0,\quad f_{xx}=-2<0. \eex</script> 于是 (2,-2)<script type="math/tex">(2,-2)</script> 为 f<script type="math/tex">f</script> 的极大值点, 且 f(2,-2)=8<script type="math/tex">f(2,-2)=8</script>.

4. 求函数 f(x,y)=x^3+y^3-3(x^2+y^2)<script type="math/tex">f(x,y)=x^3+y^3-3(x^2+y^2)</script> 的极值.

解答: 由 \bex \left\{\ba{lll} f_x=3x^2-6x=0&\ra&x=0\mbox{ or }2\\ f_y=3y^2-6y=0&\ra&y=0\mbox{ or }2 \ea\right. \eex<script type="math/tex; mode=display">\bex \left\{\ba{lll} f_x=3x^2-6x=0&\ra&x=0\mbox{ or }2\\ f_y=3y^2-6y=0&\ra&y=0\mbox{ or }2 \ea\right. \eex</script> 知 f<script type="math/tex">f</script> 的驻点有 \bex (0,0),\quad (0,2),\quad (2,0),\quad (2,2). \eex<script type="math/tex; mode=display">\bex (0,0),\quad (0,2),\quad (2,0),\quad (2,2). \eex</script> 又 \bex f_{xx}=6(x-1),\quad f_{xy}=0,\quad f_{yy}=6(y-1), \eex<script type="math/tex; mode=display">\bex f_{xx}=6(x-1),\quad f_{xy}=0,\quad f_{yy}=6(y-1), \eex</script> 我们有 \bex J(x,y)=\sev{\ba{cc} f_{xx}&f_{xy}\\ f_{yx}&f_{yy} \ea} =36(x-1)(y-1). \eex<script type="math/tex; mode=display">\bex J(x,y)=\sev{\ba{cc} f_{xx}&f_{xy}\\ f_{yx}&f_{yy} \ea} =36(x-1)(y-1). \eex</script> 于是 \item 由 J(0,0)=36>0<script type="math/tex">J(0,0)=36>0</script>, f_{xx}(0,0)=-6<0<script type="math/tex">f_{xx}(0,0)=-6<0</script> 知 f<script type="math/tex">f</script> 在 (0,0)<script type="math/tex">(0,0)</script> 处取得极大值 0<script type="math/tex">0</script>; \item 由 J(0,2)=-36<0<script type="math/tex">J(0,2)=-36<0</script> 知 (0,2)<script type="math/tex">(0,2)</script> 不是 f<script type="math/tex">f</script> 的极值; \item 由 J(2，0)=-36<0<script type="math/tex">J(2，0)=-36<0</script> 知 (2,0)<script type="math/tex">(2,0)</script> 也不是 f<script type="math/tex">f</script> 的极值; \item 由 J(2,2)=36>0<script type="math/tex">J(2,2)=36>0</script>, f_{xx}(2,2)=6>0<script type="math/tex">f_{xx}(2,2)=6>0</script> 知 f<script type="math/tex">f</script> 在 (2,2)<script type="math/tex">(2,2)</script> 处取得极小值 -8<script type="math/tex">-8</script>.

7. 求旋转抛物面 z=x^2+y^2<script type="math/tex">z=x^2+y^2</script> 与平面 x+y-z=1<script type="math/tex">x+y-z=1</script> 之间的最短距离.

解答: 旋转抛物面 z=x^2+y^2<script type="math/tex">z=x^2+y^2</script> 上一点 (x,y,z)<script type="math/tex">(x,y,z)</script> 到平面 x+y-z-1=0<script type="math/tex">x+y-z-1=0</script> 的距离为 \bex \cfrac{\sev{x+y-z-1}}{\sqrt{3}}. \eex<script type="math/tex; mode=display">\bex \cfrac{\sev{x+y-z-1}}{\sqrt{3}}. \eex</script> 于是问题转化为求解 \bex \ba{cc} \mbox{极值}&(x+y-z-1)^2\\ s.t.&x^2+y^2-z=0. \ea \eex<script type="math/tex; mode=display">\bex \ba{cc} \mbox{极值}&(x+y-z-1)^2\\ s.t.&x^2+y^2-z=0. \ea \eex</script> 记 Lagrange<script type="math/tex">Lagrange</script> 函数为 \bex L(x,y,z,\lambda) =(x+y-z-1)^2+\lambda (x^2+y^2-z). \eex<script type="math/tex; mode=display">\bex L(x,y,z,\lambda) =(x+y-z-1)^2+\lambda (x^2+y^2-z). \eex</script> 则由 \bex \left\{\ba{llll} L_x=2(x+y-z-1)+2\lambda x=0\\ L_y=2(x+y-z-1)+2\lambda y=0\\ L_z=-2(x+y-z-1)-\lambda=0\\ L_\lambda=x^2+y^2-z=0 \ea\right. \eex<script type="math/tex; mode=display">\bex \left\{\ba{llll} L_x=2(x+y-z-1)+2\lambda x=0\\ L_y=2(x+y-z-1)+2\lambda y=0\\ L_z=-2(x+y-z-1)-\lambda=0\\ L_\lambda=x^2+y^2-z=0 \ea\right. \eex</script> 知 \bex x=\cfrac{1}{2},\quad y=\cfrac{1}{2},\quad z=\cfrac{1}{2},\quad \lambda=1. \eex<script type="math/tex; mode=display">\bex x=\cfrac{1}{2},\quad y=\cfrac{1}{2},\quad z=\cfrac{1}{2},\quad \lambda=1. \eex</script> 于是所求为 \bex \cfrac{1}{\sqrt{3}}\cdot\sev{\cfrac{1}{2}+\cfrac{1}{2}-\cfrac{1}{2}-1} =\cfrac{\sqrt{3}}{6}. \eex<script type="math/tex; mode=display">\bex \cfrac{1}{\sqrt{3}}\cdot\sev{\cfrac{1}{2}+\cfrac{1}{2}-\cfrac{1}{2}-1} =\cfrac{\sqrt{3}}{6}. \eex</script>

10. 求内接于椭球面 \dps{\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}+\cfrac{z^2}{c^2}=1}<script type="math/tex">\dps{\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}+\cfrac{z^2}{c^2}=1}</script> 的长方体 (各表面平行于坐标面) 的最大体积.

解答: 以椭球面 \dps{\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}+\cfrac{z^2}{c^2}=1}<script type="math/tex">\dps{\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}+\cfrac{z^2}{c^2}=1}</script> 上一点 (x,y,z)<script type="math/tex">(x,y,z)</script> 为顶点, 各表面平行于坐标面的长方体的体积为 (2|x|)\cdot(2|y|)\cdot(2|z|)=8|xyz|<script type="math/tex">(2|x|)\cdot(2|y|)\cdot(2|z|)=8|xyz|</script>. 于是问题转化为求解 \bex \ba{cc} \mbox{极值}&x^2y^2z^2\\ s.t.& \cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}+\cfrac{z^2}{c^2}-1=0. \ea \eex<script type="math/tex; mode=display">\bex \ba{cc} \mbox{极值}&x^2y^2z^2\\ s.t.& \cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}+\cfrac{z^2}{c^2}-1=0. \ea \eex</script> 记 Lagrange<script type="math/tex">Lagrange</script> 函数为 \bex L(x,y,z,\lambda) =x^2y^2z^2+\lambda\sex{\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}+\cfrac{z^2}{c^2}-1}, \eex<script type="math/tex; mode=display">\bex L(x,y,z,\lambda) =x^2y^2z^2+\lambda\sex{\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}+\cfrac{z^2}{c^2}-1}, \eex</script> 则由 \bex \left\{\ba{llll} L_x=2xy^2z^2+\cfrac{2\lambda}{a^2}x=0,\\ L_y=2x^2yz^2+\cfrac{2\lambda}{b^2}y=0,\\ L_z=2x^2y^2z+\cfrac{2\lambda}{c^2}z=0,\\ L_\lambda=\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}+\cfrac{z^2}{c^2}-1=0 \ea\right. \eex<script type="math/tex; mode=display">\bex \left\{\ba{llll} L_x=2xy^2z^2+\cfrac{2\lambda}{a^2}x=0,\\ L_y=2x^2yz^2+\cfrac{2\lambda}{b^2}y=0,\\ L_z=2x^2y^2z+\cfrac{2\lambda}{c^2}z=0,\\ L_\lambda=\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}+\cfrac{z^2}{c^2}-1=0 \ea\right. \eex</script> 知 \bex \lambda=-\cfrac{a^2b^2c^2}{9},\quad x=\pm \cfrac{a}{\sqrt{3}},\quad y=\pm \cfrac{b}{\sqrt{3}},\quad z=\pm \cfrac{c}{\sqrt{3}}. \eex<script type="math/tex; mode=display">\bex \lambda=-\cfrac{a^2b^2c^2}{9},\quad x=\pm \cfrac{a}{\sqrt{3}},\quad y=\pm \cfrac{b}{\sqrt{3}},\quad z=\pm \cfrac{c}{\sqrt{3}}. \eex</script> 于是所求为 \bex 8\cdot \cfrac{a}{\sqrt{3}}\cdot \cfrac{b}{\sqrt{3}}\cdot \cfrac{c}{\sqrt{3}}=\cfrac{8\sqrt{3}abc}{9}. \eex<script type="math/tex; mode=display">\bex 8\cdot \cfrac{a}{\sqrt{3}}\cdot \cfrac{b}{\sqrt{3}}\cdot \cfrac{c}{\sqrt{3}}=\cfrac{8\sqrt{3}abc}{9}. \eex</script>

9 多元函数的积分学及其应用

9.1 二重积分的概念与性质

4(3). 比较二重积分 \dps{\iint_D \sin (x+y)\rd \sigma}<script type="math/tex">\dps{\iint_D \sin (x+y)\rd \sigma}</script> 与 \dps{\iint_D (x+y)\rd \sigma}<script type="math/tex">\dps{\iint_D (x+y)\rd \sigma}</script> 的大小, 其中 D<script type="math/tex">D</script> 是由直线 x+y=1,\ x+y=2<script type="math/tex">x+y=1,\ x+y=2</script> 及两坐标轴所围成的闭区域.

解答: 由于在 D<script type="math/tex">D</script> 上, 1\leq x+y\leq 2<script type="math/tex">1\leq x+y\leq 2</script>, 且 \bex & &(t-\sin t)‘=1-\cos t\geq 0\\ &\ra& t-\sin t\geq 0\quad (t\geq 0). \eex<script type="math/tex; mode=display">\bex & &(t-\sin t)‘=1-\cos t\geq 0\\ &\ra& t-\sin t\geq 0\quad (t\geq 0). \eex</script> 我们有 \bex \iint_D \sin (x+y)\rd \sigma\leq\iint_D (x+y)\rd \sigma. \eex<script type="math/tex; mode=display">\bex \iint_D \sin (x+y)\rd \sigma\leq\iint_D (x+y)\rd \sigma. \eex</script>

4(4). 比较二重积分 \dps{\iint_D \ln(x+y)\rd \sigma}<script type="math/tex">\dps{\iint_D \ln(x+y)\rd \sigma}</script> 与 \dps{\iint_D \sez{\ln(x+y)}^2\rd \sigma}<script type="math/tex">\dps{\iint_D \sez{\ln(x+y)}^2\rd \sigma}</script> 的大小, 其中 D<script type="math/tex">D</script> 是以点 (1,0)<script type="math/tex">(1,0)</script>, (1,1)<script type="math/tex">(1,1)</script> 及 (2,0)<script type="math/tex">(2,0)</script> 为顶点的三角形闭区域.

解答: 由于在 D<script type="math/tex">D</script> 上, 1\leq x+y\leq 2<script type="math/tex">1\leq x+y\leq 2</script>, 而 \bex 0\leq \ln (x+y)\leq \ln 2<1, \eex<script type="math/tex; mode=display">\bex 0\leq \ln (x+y)\leq \ln 2<1, \eex</script> \bex \ln(x+y)\geq\sez{\ln (x+y)}^2\quad ((x,y)\in D). \eex<script type="math/tex; mode=display">\bex \ln(x+y)\geq\sez{\ln (x+y)}^2\quad ((x,y)\in D). \eex</script> 于是 \bex \iint_D \ln(x+y)\rd \sigma \geq \iint_D \sez{\ln(x+y)}^2\rd \sigma. \eex<script type="math/tex; mode=display">\bex \iint_D \ln(x+y)\rd \sigma \geq \iint_D \sez{\ln(x+y)}^2\rd \sigma. \eex</script>

5(3). 估计二重积分 \dps{I=\iint_D (x+y+10)\rd \sigma}<script type="math/tex">\dps{I=\iint_D (x+y+10)\rd \sigma}</script> 的值, 其中 \bex D=\sed{(x,y);\ 0\leq x\leq 3,\ 0\leq y\leq 1}. \eex<script type="math/tex; mode=display">\bex D=\sed{(x,y);\ 0\leq x\leq 3,\ 0\leq y\leq 1}. \eex</script>

解答: 由 0\leq x\leq 3<script type="math/tex">0\leq x\leq 3</script>, 0\leq y\leq 1<script type="math/tex">0\leq y\leq 1</script> 知 \bex 10\leq x+y+10\leq 14. \eex<script type="math/tex; mode=display">\bex 10\leq x+y+10\leq 14. \eex</script> 又 D<script type="math/tex">D</script> 的面积为 3\times 1=3<script type="math/tex">3\times 1=3</script>, 我们有 \bex 30=3\cdot 10\leq I=\iint_D (x+y+10)\rd \sigma \leq 3\cdot 14=42. \eex<script type="math/tex; mode=display">\bex 30=3\cdot 10\leq I=\iint_D (x+y+10)\rd \sigma \leq 3\cdot 14=42. \eex</script>

5(4). 估计二重积分 \dps{I=\iint_D (x^2+4y^2+9)\rd \sigma}<script type="math/tex">\dps{I=\iint_D (x^2+4y^2+9)\rd \sigma}</script> 的值, 其中 \bex D=\sed{(x,y);\ x^2+y^2\leq 1}. \eex<script type="math/tex; mode=display">\bex D=\sed{(x,y);\ x^2+y^2\leq 1}. \eex</script>

解答: 由于在 D<script type="math/tex">D</script> 上, x^2+y^2\leq 1<script type="math/tex">x^2+y^2\leq 1</script>, 可令 \bex x=r\cos\theta,\quad y=r\sin\theta, \eex<script type="math/tex; mode=display">\bex x=r\cos\theta,\quad y=r\sin\theta, \eex</script> 其中 0\leq r\leq 1<script type="math/tex">0\leq r\leq 1</script>, 0\leq \theta\leq 2\pi<script type="math/tex">0\leq \theta\leq 2\pi</script>. 于是 \bex x^2+4y^2+9 =r^2(\cos^2\theta+4\sin^2\theta)+9 =r^2(1+3\sin^2\theta)+9 \eex<script type="math/tex; mode=display">\bex x^2+4y^2+9 =r^2(\cos^2\theta+4\sin^2\theta)+9 =r^2(1+3\sin^2\theta)+9 \eex</script> 在 9<script type="math/tex">9</script> 与 13<script type="math/tex">13</script> 之间. 故而 \bex 9\pi \leq I=\iint_D (x^2+4y^2+9)\rd \sigma \leq 13\pi. \eex<script type="math/tex; mode=display">\bex 9\pi \leq I=\iint_D (x^2+4y^2+9)\rd \sigma \leq 13\pi. \eex</script>

9.2 二重积分的计算法

1(3). 画出下列积分区域 \dps{D=\sed{(x,y);\ 0\leq x\leq 1,\ 0\leq y\leq 1}}<script type="math/tex">\dps{D=\sed{(x,y);\ 0\leq x\leq 1,\ 0\leq y\leq 1}}</script> 的图形, 并且计算二重积分 \dps{\iint_D \sqrt{x+y}\rd \sigma}<script type="math/tex">\dps{\iint_D \sqrt{x+y}\rd \sigma}</script>.

解答: 积分区域为

\bex \iint_D \sqrt{x+y}\rd \sigma &=&\int_0^1 \rd x\int_0^1 \sqrt{x+y}\rd y\\ &=&\int_0^1 \sez{\left.\cfrac{2}{3}(x+y)^\cfrac{3}{2}\right|^{y=1}_{y=0}}\rd x\\ &=&\int_0^1 \sez{\cfrac{2}{3}(x+1)^\cfrac{3}{2} -\cfrac{2}{3}x^\cfrac{3}{2}}\rd x\\ &=&\left.\sez{\cfrac{4}{15}(x+1)^\cfrac{5}{2} -\cfrac{4}{15}x^\cfrac{5}{2}}\right|^{x=1}_{x=0}\\ &=&\cfrac{4}{15}(4\sqrt{2}-1)-\cfrac{4}{15}\\ &=&\cfrac{8(2\sqrt{2}-1)}{15}. \eex<script type="math/tex; mode=display">\bex \iint_D \sqrt{x+y}\rd \sigma &=&\int_0^1 \rd x\int_0^1 \sqrt{x+y}\rd y\\ &=&\int_0^1 \sez{\left.\cfrac{2}{3}(x+y)^\cfrac{3}{2}\right|^{y=1}_{y=0}}\rd x\\ &=&\int_0^1 \sez{\cfrac{2}{3}(x+1)^\cfrac{3}{2} -\cfrac{2}{3}x^\cfrac{3}{2}}\rd x\\ &=&\left.\sez{\cfrac{4}{15}(x+1)^\cfrac{5}{2} -\cfrac{4}{15}x^\cfrac{5}{2}}\right|^{x=1}_{x=0}\\ &=&\cfrac{4}{15}(4\sqrt{2}-1)-\cfrac{4}{15}\\ &=&\cfrac{8(2\sqrt{2}-1)}{15}. \eex</script>

1(9). 设 D<script type="math/tex">D</script> 是由抛物线 y=x^2<script type="math/tex">y=x^2</script>, 直线 x+2y-3=0<script type="math/tex">x+2y-3=0</script> 及 x<script type="math/tex">x</script> 轴围成的闭区域, 试画出 D<script type="math/tex">D</script> 的图形, 并且计算二重积分 \dps{\iint_D xy\rd \sigma}<script type="math/tex">\dps{\iint_D xy\rd \sigma}</script>.

解答: D<script type="math/tex">D</script> 的图形为

\bex \iint_D xy\rd \sigma &=&\int_0^1 \rd y \int_{\sqrt{y}}^{3-2y} xy\rd x\\ &=&\int_0^1 \sez{\left.\cfrac{x^2y}{2}\right|^{x=3-2y}_{x=\sqrt{y}}}\rd y\\ &=&\cfrac{1}{2}\int_0^1 \sez{y(2y-3)^2-y^2}\rd y\\ &=&\cfrac{1}{2}\int_0^1 \sex{4y^3-13y^2+9y}\rd y\\ &=&\cfrac{1}{2}\left.\sex{y^4-\cfrac{13}{3}y^3+\cfrac{9}{2}y^2}\right|^{x=1}_{x=0}\\ &=&\cfrac{1}{2}\sex{1-\cfrac{13}{3}+\cfrac{9}{2}}\\ &=&\cfrac{7}{12}. \eex<script type="math/tex; mode=display">\bex \iint_D xy\rd \sigma &=&\int_0^1 \rd y \int_{\sqrt{y}}^{3-2y} xy\rd x\\ &=&\int_0^1 \sez{\left.\cfrac{x^2y}{2}\right|^{x=3-2y}_{x=\sqrt{y}}}\rd y\\ &=&\cfrac{1}{2}\int_0^1 \sez{y(2y-3)^2-y^2}\rd y\\ &=&\cfrac{1}{2}\int_0^1 \sex{4y^3-13y^2+9y}\rd y\\ &=&\cfrac{1}{2}\left.\sex{y^4-\cfrac{13}{3}y^3+\cfrac{9}{2}y^2}\right|^{x=1}_{x=0}\\ &=&\cfrac{1}{2}\sex{1-\cfrac{13}{3}+\cfrac{9}{2}}\\ &=&\cfrac{7}{12}. \eex</script>

3(3). 化二重积分 \dps{I=\iint_D f(x,y)\rd \sigma}<script type="math/tex">\dps{I=\iint_D f(x,y)\rd \sigma}</script> 为二次积分 (分别列出对两个变量先后次序不同的两个二次积分), 其中 f(x,y)<script type="math/tex">f(x,y)</script> 在 D<script type="math/tex">D</script> 上连续, 积分区域 D<script type="math/tex">D</script> 是由抛物线 y=x^2<script type="math/tex">y=x^2</script> 与圆周 x^2+y^2=2<script type="math/tex">x^2+y^2=2</script> 所围成的位于 x<script type="math/tex">x</script> 轴上方的闭区域.

解答: D<script type="math/tex">D</script> 为

\bex I&=&\iint_D f(x,y)\rd \sigma\\ &=&\int_{-1}^1\rd x\int_{x^2}^{\sqrt{2-x^2}}f(x,y)\rd y\\ &=&\int_0^1 \rd y\int_{-\sqrt{y}}^{\sqrt{y}} f(x,y)\rd x +\int_1^{\sqrt{2}}\rd y \int_{-\sqrt{2-y^2}}^{\sqrt{2-y^2}} f(x,y)\rd x. \eex<script type="math/tex; mode=display">\bex I&=&\iint_D f(x,y)\rd \sigma\\ &=&\int_{-1}^1\rd x\int_{x^2}^{\sqrt{2-x^2}}f(x,y)\rd y\\ &=&\int_0^1 \rd y\int_{-\sqrt{y}}^{\sqrt{y}} f(x,y)\rd x +\int_1^{\sqrt{2}}\rd y \int_{-\sqrt{2-y^2}}^{\sqrt{2-y^2}} f(x,y)\rd x. \eex</script>

5. 设平面薄片占据的区域 D<script type="math/tex">D</script> 由直线 x+y=3<script type="math/tex">x+y=3</script>, y=x-3<script type="math/tex">y=x-3</script> 与 x<script type="math/tex">x</script> 轴围成, 它的面密度 \mu(x,y)=x^2+y^2<script type="math/tex">\mu(x,y)=x^2+y^2</script>, 求该薄片的质量.

解答: D<script type="math/tex">D</script> 为 9_2_5

\bex m&=&\iint_D \mu(x,y)\rd \sigma\\ &=&\int_0^1 \rd y\int_{y+1}^{3-y}(x^2+y^2)\rd x\\ &=&\int_0^1 \sez{\left.\cfrac{x^3}{3}+y^2x\right|^{x=3-y}_{x=y+1}}\rd y\\ &=&\int_0^1 \sez{ -\cfrac{(y-3)^3}{3} +3y^2 -y^3 -\cfrac{(y+1)^3}{3} -y^3 -y^2 }\rd y\\ &=&\left.\sez{ -\cfrac{(y-3)^4}{12} -\cfrac{(y+1)^4}{12} +\cfrac{2y^3}{3} -\cfrac{y^4}{2} }\right|^{y=1}_{y=0}\\ &=&\cfrac{65}{12} -\cfrac{15}{12} +\cfrac{2}{3} -\cfrac{1}{2}\\ &=&\cfrac{13}{3}. \eex<script type="math/tex; mode=display">\bex m&=&\iint_D \mu(x,y)\rd \sigma\\ &=&\int_0^1 \rd y\int_{y+1}^{3-y}(x^2+y^2)\rd x\\ &=&\int_0^1 \sez{\left.\cfrac{x^3}{3}+y^2x\right|^{x=3-y}_{x=y+1}}\rd y\\ &=&\int_0^1 \sez{ -\cfrac{(y-3)^3}{3} +3y^2 -y^3 -\cfrac{(y+1)^3}{3} -y^3 -y^2 }\rd y\\ &=&\left.\sez{ -\cfrac{(y-3)^4}{12} -\cfrac{(y+1)^4}{12} +\cfrac{2y^3}{3} -\cfrac{y^4}{2} }\right|^{y=1}_{y=0}\\ &=&\cfrac{65}{12} -\cfrac{15}{12} +\cfrac{2}{3} -\cfrac{1}{2}\\ &=&\cfrac{13}{3}. \eex</script>

10(2). 设 D<script type="math/tex">D</script> 是由圆周 x^2+y^2=2y<script type="math/tex">x^2+y^2=2y</script> 与 y-<script type="math/tex">y-</script>轴所围成的位于第一象限内的闭区域, 求 \dps{\iint_D \sqrt{x^2+y^2}\rd \sigma}<script type="math/tex">\dps{\iint_D \sqrt{x^2+y^2}\rd \sigma}</script>.

解答: D<script type="math/tex">D</script> 为

\bex \iint_D \sqrt{x^2+y^2}\rd \sigma &=&\int_0^\cfrac{\pi}{2}\rd \theta \int_0^{2\sin\theta} \rho\cdot\rho \rd \rho\\ &=&\int_0^\cfrac{\pi}{2}\cfrac{8\sin^3\theta}{3}\rd \theta\\ &=&\cfrac{8}{3}\cdot\cfrac{2}{3}\\ & &\sex{\int \sin^3\theta\rd \theta =\int \sin\theta\sex{1-\cos^2\theta}\rd\theta \atop =-\cos\theta+\cfrac{\cos^3\theta}{3}+C}\\ &=&\cfrac{16}{9}. \eex<script type="math/tex; mode=display">\bex \iint_D \sqrt{x^2+y^2}\rd \sigma &=&\int_0^\cfrac{\pi}{2}\rd \theta \int_0^{2\sin\theta} \rho\cdot\rho \rd \rho\\ &=&\int_0^\cfrac{\pi}{2}\cfrac{8\sin^3\theta}{3}\rd \theta\\ &=&\cfrac{8}{3}\cdot\cfrac{2}{3}\\ & &\sex{\int \sin^3\theta\rd \theta =\int \sin\theta\sex{1-\cos^2\theta}\rd\theta \atop =-\cos\theta+\cfrac{\cos^3\theta}{3}+C}\\ &=&\cfrac{16}{9}. \eex</script>

12. 设平面薄片占据的闭区域 D<script type="math/tex">D</script> 是由螺线 \rho=2\theta<script type="math/tex">\rho=2\theta</script> 的一段弧 \dps{\sex{0\leq \theta\leq\cfrac{\pi}{2}}}<script type="math/tex">\dps{\sex{0\leq \theta\leq\cfrac{\pi}{2}}}</script> 与射线 \dps{\theta=\cfrac{\pi}{2}}<script type="math/tex">\dps{\theta=\cfrac{\pi}{2}}</script> 所围成, 它的面密度 \mu(x,y)=\sqrt{x^2+y^2}<script type="math/tex">\mu(x,y)=\sqrt{x^2+y^2}</script>, 求该薄片的质量.

解答: D<script type="math/tex">D</script> 为

\bex m&=&\iint_D \mu(x,y)\rd \sigma\\ &=&\int_0^\cfrac{\pi}{2} \rd \theta \int_0^{2\theta}\rho\cdot\rho \rd \rho\\ &=&\int_0^\cfrac{\pi}{2} \cfrac{8\theta^3}{3}\rd \theta\\ &=&\cfrac{2}{3}\cdot\sex{\cfrac{\pi}{2}}^4\\ &=&\cfrac{\pi^4}{24}. \eex<script type="math/tex; mode=display">\bex m&=&\iint_D \mu(x,y)\rd \sigma\\ &=&\int_0^\cfrac{\pi}{2} \rd \theta \int_0^{2\theta}\rho\cdot\rho \rd \rho\\ &=&\int_0^\cfrac{\pi}{2} \cfrac{8\theta^3}{3}\rd \theta\\ &=&\cfrac{2}{3}\cdot\sex{\cfrac{\pi}{2}}^4\\ &=&\cfrac{\pi^4}{24}. \eex</script>

13(1). 计算心形线 \rho=1+\cos\theta<script type="math/tex">\rho=1+\cos\theta</script> 所围成平面图形的面积.

解答:图为

\bex \mbox{Area} &=&\iint_D \rd \sigma\\ &=&\int_0^{2\pi}\rd\theta \int_0^{1+\cos\theta} \rho \rd \rho\\ &=&\int_0^{2\pi} \cfrac{(1+\cos\theta)^2}{2}\rd \theta\\ &=&\int_0^{2\pi} \cfrac{1+2\cos\theta +\cfrac{1+\cos2\theta}{2}}{2} \rd \theta\\ &=&\int_0^{2\pi}\cfrac{3}{4}\rd\theta\\ &=&\cfrac{3\pi}{2}. \eex<script type="math/tex; mode=display">\bex \mbox{Area} &=&\iint_D \rd \sigma\\ &=&\int_0^{2\pi}\rd\theta \int_0^{1+\cos\theta} \rho \rd \rho\\ &=&\int_0^{2\pi} \cfrac{(1+\cos\theta)^2}{2}\rd \theta\\ &=&\int_0^{2\pi} \cfrac{1+2\cos\theta +\cfrac{1+\cos2\theta}{2}}{2} \rd \theta\\ &=&\int_0^{2\pi}\cfrac{3}{4}\rd\theta\\ &=&\cfrac{3\pi}{2}. \eex</script>

9.3 二重积分的应用

2. 求圆锥面 z=\sqrt{x^2+y^2}<script type="math/tex">z=\sqrt{x^2+y^2}</script> 被柱面 z^2=2x<script type="math/tex">z^2=2x</script> 所割下的那部分曲面的面积.

解答: 曲面为

由 \bex \left.\ba{ll} x^2+y^2=z^2\\ z^2=2x \ea\right\} \ra (x-1)^2+y^2=1\quad \sex{\rho=2\cos\theta} \eex<script type="math/tex; mode=display">\bex \left.\ba{ll} x^2+y^2=z^2\\ z^2=2x \ea\right\} \ra (x-1)^2+y^2=1\quad \sex{\rho=2\cos\theta} \eex</script> 及该曲面在 x<script type="math/tex">x</script>-o<script type="math/tex">o</script>-y<script type="math/tex">y</script> 面上的投影

\bex S&=&\iint_{(x-1)^2+y^2\leq 1} \sqrt{1+\sex{\cfrac{2x}{2\sqrt{x^2+y^2}}}^2 +\sex{\cfrac{2y}{2\sqrt{x^2+y^2}}}^2}\rd x\rd y\\ &=&\sqrt{2}\iint_{(x-1)^2+y^2\leq 1}\rd x\rd y\\ &=&\sqrt{2}\pi. \eex<script type="math/tex; mode=display">\bex S&=&\iint_{(x-1)^2+y^2\leq 1} \sqrt{1+\sex{\cfrac{2x}{2\sqrt{x^2+y^2}}}^2 +\sex{\cfrac{2y}{2\sqrt{x^2+y^2}}}^2}\rd x\rd y\\ &=&\sqrt{2}\iint_{(x-1)^2+y^2\leq 1}\rd x\rd y\\ &=&\sqrt{2}\pi. \eex</script>

4(1). 设 D<script type="math/tex">D</script> 是由抛物线 y=\sqrt{2x}<script type="math/tex">y=\sqrt{2x}</script> 与直线 x=1,\ y=0<script type="math/tex">x=1,\ y=0</script> 所围成的闭区域, 求 D<script type="math/tex">D</script> 的质心.

解答:图为

由 \bex \iint_D \rd \sigma =\int_0^1 \rd x \int_0^{\sqrt{2x}}\rd y =\int_0^1 \sqrt{2x}\rd x =\cfrac{2\sqrt{2}}{3}; \eex<script type="math/tex; mode=display">\bex \iint_D \rd \sigma =\int_0^1 \rd x \int_0^{\sqrt{2x}}\rd y =\int_0^1 \sqrt{2x}\rd x =\cfrac{2\sqrt{2}}{3}; \eex</script> \bex \iint_D \rd \sigma =\int_0^1 x\rd \int_0^{\sqrt{2x}} \rd y =\int_0^1 \sqrt{2}x^\cfrac{3}{2}\rd x =\cfrac{2\sqrt{2}}{3}; \eex<script type="math/tex; mode=display">\bex \iint_D \rd \sigma =\int_0^1 x\rd \int_0^{\sqrt{2x}} \rd y =\int_0^1 \sqrt{2}x^\cfrac{3}{2}\rd x =\cfrac{2\sqrt{2}}{3}; \eex</script> \bex \iint_D y\rd y =\int_0^1 \rd x \int_0^{\sqrt{2x}}y\rd y =\int_0^1 x\rd x =\cfrac{1}{2} \eex<script type="math/tex; mode=display">\bex \iint_D y\rd y =\int_0^1 \rd x \int_0^{\sqrt{2x}}y\rd y =\int_0^1 x\rd x =\cfrac{1}{2} \eex</script> 知 \bex \bar x =\cfrac{\iint_D x\rd x} {\iint_D\rd x} =\cfrac{3}{5},\quad \bar y =\cfrac{\iint_D y\rd x} {\iint_D\rd x} =\cfrac{3\sqrt{2}}{8}. \eex<script type="math/tex; mode=display">\bex \bar x =\cfrac{\iint_D x\rd x} {\iint_D\rd x} =\cfrac{3}{5},\quad \bar y =\cfrac{\iint_D y\rd x} {\iint_D\rd x} =\cfrac{3\sqrt{2}}{8}. \eex</script> 于是 D<script type="math/tex">D</script> 的质心为 \dps{\sex{\cfrac{3}{5},\cfrac{3\sqrt{2}}{8}}}<script type="math/tex">\dps{\sex{\cfrac{3}{5},\cfrac{3\sqrt{2}}{8}}}</script>.

8(2). 设均匀薄片 (面密度为常数 \mu<script type="math/tex">\mu</script>) 占据 \bex D=\sed{(x,y);\ 0\leq x\leq a,\ 0\leq y\leq b}, \eex<script type="math/tex; mode=display">\bex D=\sed{(x,y);\ 0\leq x\leq a,\ 0\leq y\leq b}, \eex</script> 求它分别对于 x<script type="math/tex">x</script> 轴, y<script type="math/tex">y</script> 轴 与直线 \dps{l:\ y=\cfrac{b}{a}x}<script type="math/tex">\dps{l:\ y=\cfrac{b}{a}x}</script> 的转动惯量 I_x<script type="math/tex">I_x</script>, I_y<script type="math/tex">I_y</script> 与 I_l<script type="math/tex">I_l</script>.

解答: \bex I_x=\iint_D y^2\mu\rd \sigma =\mu\int_0^a\rd x \int_0^b y^2\rd y =\cfrac{ab^3\mu}{3}; \eex<script type="math/tex; mode=display">\bex I_x=\iint_D y^2\mu\rd \sigma =\mu\int_0^a\rd x \int_0^b y^2\rd y =\cfrac{ab^3\mu}{3}; \eex</script> \bex I_y=\iint_D x^2\mu \rd \sigma =\mu\int_0^a x^2\rd x \int_0^b \rd y =\cfrac{a^3b\mu}{3}; \eex<script type="math/tex; mode=display">\bex I_y=\iint_D x^2\mu \rd \sigma =\mu\int_0^a x^2\rd x \int_0^b \rd y =\cfrac{a^3b\mu}{3}; \eex</script> \bex I_l&=&\iint_D \cfrac{(bx-ay)^2}{a^2+b^2}\mu\rd \sigma\\ &=&\cfrac{b^2}{a^2+b^2}I_y -\cfrac{2ab\mu}{a^2+b^2} \iint_D xy\rd \sigma +\cfrac{a^2}{a^2+b^2}I_x\\ &=&\cfrac{b^2}{a^2+b^2}\cdot \cfrac{a^3b\mu}{3} -\cfrac{2ab\mu}{a^2+b^2} \int_0^ax\rd x\cdot\int_0^b y\rd y +\cfrac{a^2}{a^2+b^2}\cdot\cfrac{ab^3\mu}{3}\\ &=&\cfrac{a^3b^3\mu}{6(a^2+b^2)}. \eex<script type="math/tex; mode=display">\bex I_l&=&\iint_D \cfrac{(bx-ay)^2}{a^2+b^2}\mu\rd \sigma\\ &=&\cfrac{b^2}{a^2+b^2}I_y -\cfrac{2ab\mu}{a^2+b^2} \iint_D xy\rd \sigma +\cfrac{a^2}{a^2+b^2}I_x\\ &=&\cfrac{b^2}{a^2+b^2}\cdot \cfrac{a^3b\mu}{3} -\cfrac{2ab\mu}{a^2+b^2} \int_0^ax\rd x\cdot\int_0^b y\rd y +\cfrac{a^2}{a^2+b^2}\cdot\cfrac{ab^3\mu}{3}\\ &=&\cfrac{a^3b^3\mu}{6(a^2+b^2)}. \eex</script>

9.4 三重积分

3(1). 计算 \dps{\iint_\Omega xy\rd v}<script type="math/tex">\dps{\iint_\Omega xy\rd v}</script>, 其中 \Omega<script type="math/tex">\Omega</script> 是由三个坐标面与平面 \dps{x+\cfrac{y}{2}+\cfrac{z}{3}=1}<script type="math/tex">\dps{x+\cfrac{y}{2}+\cfrac{z}{3}=1}</script> 所围成的闭区域.

解答: \bex \iint_\Omega xy\rd v &=&\int_0^1 \rd x\int_0^{2(1-x)}\rd y \int_0^{3\sex{1-x-\cfrac{y}{2}}}xy\rd z\\ &=&3\int_0^1 x\rd x\int_0^{2(1-x)}y\sex{1-x-\cfrac{y}{2}}\rd y\\ &=&2\int_0^1 x(1-x)^3\rd x\\ &=&2\int_0^1 \sex{x-3x^2+3x^3-x^4}\rd x\\ &=&2\sex{\cfrac{1}{2}-1+\cfrac{3}{4}-\cfrac{1}{5}}\\ &=&\cfrac{1}{10}. \eex<script type="math/tex; mode=display">\bex \iint_\Omega xy\rd v &=&\int_0^1 \rd x\int_0^{2(1-x)}\rd y \int_0^{3\sex{1-x-\cfrac{y}{2}}}xy\rd z\\ &=&3\int_0^1 x\rd x\int_0^{2(1-x)}y\sex{1-x-\cfrac{y}{2}}\rd y\\ &=&2\int_0^1 x(1-x)^3\rd x\\ &=&2\int_0^1 \sex{x-3x^2+3x^3-x^4}\rd x\\ &=&2\sex{\cfrac{1}{2}-1+\cfrac{3}{4}-\cfrac{1}{5}}\\ &=&\cfrac{1}{10}. \eex</script>

3(5). 计算 \dps{\iint_\Omega z^2\rd v}<script type="math/tex">\dps{\iint_\Omega z^2\rd v}</script>, 其中 \Omega<script type="math/tex">\Omega</script> 是由球面 x^2+y^2+z^2=2z<script type="math/tex">x^2+y^2+z^2=2z</script> 所围成的闭区域.

解答: \bex \iint_\Omega z^2\rd v &=&\int_0^2\rd z\iint_{x^2+y^2\leq 2z-z^2}z^2\rd x\rd y\\ &=&\int_0^2z^2\cdot \pi (2z-z^2)\rd z\\ &=&\cfrac{8\pi}{5}. \eex<script type="math/tex; mode=display">\bex \iint_\Omega z^2\rd v &=&\int_0^2\rd z\iint_{x^2+y^2\leq 2z-z^2}z^2\rd x\rd y\\ &=&\int_0^2z^2\cdot \pi (2z-z^2)\rd z\\ &=&\cfrac{8\pi}{5}. \eex</script>

4(2). 计算 \dps{\iint_\Omega z\sqrt{x^2+y^2}\rd v}<script type="math/tex">\dps{\iint_\Omega z\sqrt{x^2+y^2}\rd v}</script>, 其中 \Omega<script type="math/tex">\Omega</script> 是由旋转抛物面 z=x^2+y^2<script type="math/tex">z=x^2+y^2</script> 与平面 z=1<script type="math/tex">z=1</script> 所围成的闭区域.

解答: \bex \iint_\Omega z\sqrt{x^2+y^2}\rd v &=&\int_0^1 z\rd z \iint_{x^2+y^2\leq z}\sqrt{x^2+y^2}\rd x\rd y\\ &=&\int_0^1 z\rd z \int_0^{2\pi}\rd \theta \int_0^{\sqrt{z}}r\cdot r\rd r\\ &=&\cfrac{2\pi}{3} \int_0^1 z^\cfrac{5}{2}\rd z\\ &=&\cfrac{4\pi}{21}. \eex<script type="math/tex; mode=display">\bex \iint_\Omega z\sqrt{x^2+y^2}\rd v &=&\int_0^1 z\rd z \iint_{x^2+y^2\leq z}\sqrt{x^2+y^2}\rd x\rd y\\ &=&\int_0^1 z\rd z \int_0^{2\pi}\rd \theta \int_0^{\sqrt{z}}r\cdot r\rd r\\ &=&\cfrac{2\pi}{3} \int_0^1 z^\cfrac{5}{2}\rd z\\ &=&\cfrac{4\pi}{21}. \eex</script>

5. 设密度为常量 \mu<script type="math/tex">\mu</script> 的匀质物体占据由抛物面 z=3-x^2-y^2<script type="math/tex">z=3-x^2-y^2</script> 与平面 \sev{x}=1,\ \sev{y}=1,\ z=0<script type="math/tex">\sev{x}=1,\ \sev{y}=1,\ z=0</script> 所围成的闭区域. 试求: \item 物体的质量; \item 物体的质心; \item 物体对于 z<script type="math/tex">z</script> 轴的转动惯量.

解答: \item \bex M&=&\iint_\Omega \mu\rd x\rd y\rd z\\ &=&\mu \int_{-1}^1\rd x \int_{-1}^1 \rd y\int_0^{3-x^2-y^2}\rd z\\ &=&\cfrac{28\mu}{3}. \eex<script type="math/tex; mode=display">\bex M&=&\iint_\Omega \mu\rd x\rd y\rd z\\ &=&\mu \int_{-1}^1\rd x \int_{-1}^1 \rd y\int_0^{3-x^2-y^2}\rd z\\ &=&\cfrac{28\mu}{3}. \eex</script> \item 由于物体是均匀的, 且关于 y<script type="math/tex">y</script>-o<script type="math/tex">o</script>-z<script type="math/tex">z</script>, z<script type="math/tex">z</script>-o<script type="math/tex">o</script>-x<script type="math/tex">x</script> 面对称, 故 \bex \bar x=\bar y=0. \eex<script type="math/tex; mode=display">\bex \bar x=\bar y=0. \eex</script> 往求 \bar z<script type="math/tex">\bar z</script>. \bex \bar z &=&\cfrac{1}{M}\iint_\Omega z \mu\rd x\rd y\rd z\\ &=&\cfrac{3}{28}\iint_\Omega z\rd x\rd y\rd z\\ &=&\cfrac{3}{28}\int_{-1}^1\rd x \int_{-1}^1 \rd y \int_0^{3-x^2-y^2}z\rd z\\ &=&\cfrac{253}{210}. \eex<script type="math/tex; mode=display">\bex \bar z &=&\cfrac{1}{M}\iint_\Omega z \mu\rd x\rd y\rd z\\ &=&\cfrac{3}{28}\iint_\Omega z\rd x\rd y\rd z\\ &=&\cfrac{3}{28}\int_{-1}^1\rd x \int_{-1}^1 \rd y \int_0^{3-x^2-y^2}z\rd z\\ &=&\cfrac{253}{210}. \eex</script> 于是质心为 \dps{\sex{0,0,\cfrac{253}{210}}}<script type="math/tex">\dps{\sex{0,0,\cfrac{253}{210}}}</script>. \item \bex I_z&=&\iint_\Omega (x^2+y^2)\mu\rd x\rd y\rd z\\ &=&2\mu \iint_\Omega x^2\mu\rd x\rd y\rd z\\ &=&2\mu\int_{-1}^1 x^2\rd x \int_{-1}^1 \rd y \int_0^{3-x^2-y^2}\rd z\\ &=&\cfrac{248\mu}{45}. \eex<script type="math/tex; mode=display">\bex I_z&=&\iint_\Omega (x^2+y^2)\mu\rd x\rd y\rd z\\ &=&2\mu \iint_\Omega x^2\mu\rd x\rd y\rd z\\ &=&2\mu\int_{-1}^1 x^2\rd x \int_{-1}^1 \rd y \int_0^{3-x^2-y^2}\rd z\\ &=&\cfrac{248\mu}{45}. \eex</script>

9.5 曲线积分

1(1). 计算 \dps{\oint_L(x^2+y^2)^n\rd s}<script type="math/tex">\dps{\oint_L(x^2+y^2)^n\rd s}</script>, 其中 L<script type="math/tex">L</script> 为圆周 \dps{x^2+y^2=a^2}<script type="math/tex">\dps{x^2+y^2=a^2}</script>.

解答: 令 \bex \left\{\ba{ll} x=a\cos\theta\\ y=a\sin\theta \ea\right. \quad\sex{0\leq \theta\leq 2\pi}. \eex<script type="math/tex; mode=display">\bex \left\{\ba{ll} x=a\cos\theta\\ y=a\sin\theta \ea\right. \quad\sex{0\leq \theta\leq 2\pi}. \eex</script> 则有 \bex \sqrt{x‘^2(\theta)+y‘^2(\theta)}=a, \eex<script type="math/tex; mode=display">\bex \sqrt{x‘^2(\theta)+y‘^2(\theta)}=a, \eex</script> 于是 \bex \oint_L(x^2+y^2)^n\rd s =\int_0^{2\pi}a^{2n}\cdot a\rd \theta =2\pi a^{2n+1}. \eex<script type="math/tex; mode=display">\bex \oint_L(x^2+y^2)^n\rd s =\int_0^{2\pi}a^{2n}\cdot a\rd \theta =2\pi a^{2n+1}. \eex</script>

1(3). 计算 \dps{\int_L y\rd s}<script type="math/tex">\dps{\int_L y\rd s}</script>, 其中 L<script type="math/tex">L</script> 为抛物线 \dps{y^2=4x}<script type="math/tex">\dps{y^2=4x}</script> 上点 与点 间的一段弧.

解答: L<script type="math/tex">L</script>　为 \bex x=\cfrac{y^2}{4}\quad\sex{0\leq y\leq 2}. \eex<script type="math/tex; mode=display">\bex x=\cfrac{y^2}{4}\quad\sex{0\leq y\leq 2}. \eex</script> 由 \bex \sqrt{1+x‘^2(y)}=\sqrt{1+\cfrac{y^2}{4}} \eex<script type="math/tex; mode=display">\bex \sqrt{1+x‘^2(y)}=\sqrt{1+\cfrac{y^2}{4}} \eex</script> 知 \bex \int_L y\rd s &=&\int_0^2y\sqrt{1+\cfrac{y^2}{4}}\rd y\\ &=&\cfrac{4}{3}\left.\sex{1+\cfrac{y^2}{4}}^\cfrac{3}{2}\right|^2_0\\ &=&\cfrac{4}{3}\sex{2\sqrt{2}-1}. \eex<script type="math/tex; mode=display">\bex \int_L y\rd s &=&\int_0^2y\sqrt{1+\cfrac{y^2}{4}}\rd y\\ &=&\cfrac{4}{3}\left.\sex{1+\cfrac{y^2}{4}}^\cfrac{3}{2}\right|^2_0\\ &=&\cfrac{4}{3}\sex{2\sqrt{2}-1}. \eex</script>

1(4). 计算 \dps{\int_L (x+y)\rd s}<script type="math/tex">\dps{\int_L (x+y)\rd s}</script>, 其中 L<script type="math/tex">L</script> 为连接点 \dps{(1,0)}<script type="math/tex">\dps{(1,0)}</script> 与 (0,1)<script type="math/tex">(0,1)</script> 的直线段.

解答: L<script type="math/tex">L</script> 为 \bex y=1-x\quad\sex{0\leq x\leq 1}. \eex<script type="math/tex; mode=display">\bex y=1-x\quad\sex{0\leq x\leq 1}. \eex</script> 由 \bex \sqrt{1+y‘^2(x)}=\sqrt{2} \eex<script type="math/tex; mode=display">\bex \sqrt{1+y‘^2(x)}=\sqrt{2} \eex</script> 知 \bex \int_L (x+y)\rd s=\int_0^1 \sez{x+(1-x)}\cdot\sqrt{2}\rd x =\sqrt{2}. \eex<script type="math/tex; mode=display">\bex \int_L (x+y)\rd s=\int_0^1 \sez{x+(1-x)}\cdot\sqrt{2}\rd x =\sqrt{2}. \eex</script>

1(10). 计算 \dps{\oint_L \sqrt{x^2+y^2}\rd s}<script type="math/tex">\dps{\oint_L \sqrt{x^2+y^2}\rd s}</script>, 其中 L<script type="math/tex">L</script> 为上半圆周 \dps{x^2+y^2=2x\ (y\geq0)}<script type="math/tex">\dps{x^2+y^2=2x\ (y\geq0)}</script> 与 x<script type="math/tex">x</script> 轴所围成的区域的整个边界.

解答: \bex \oint_L \sqrt{x^2+y^2}\rd s &=&\int_0^2 x\rd x +\int_0^\cfrac{\pi}{2}2\cos\theta\cdot \sqrt{\sex{2\cos\theta\sin\theta}‘^2+\sex{2\cos\theta\sin\theta}‘^2}\rd \theta\\ &=&2+4\int_0^\cfrac{\pi}{2}\cos\theta\rd \theta\\ &=&6. \eex<script type="math/tex; mode=display">\bex \oint_L \sqrt{x^2+y^2}\rd s &=&\int_0^2 x\rd x +\int_0^\cfrac{\pi}{2}2\cos\theta\cdot \sqrt{\sex{2\cos\theta\sin\theta}‘^2+\sex{2\cos\theta\sin\theta}‘^2}\rd \theta\\ &=&2+4\int_0^\cfrac{\pi}{2}\cos\theta\rd \theta\\ &=&6. \eex</script>

3(1). 计算 \dps{\int_L(x^2-y^2)\rd x}<script type="math/tex">\dps{\int_L(x^2-y^2)\rd x}</script>, 其中 L<script type="math/tex">L</script> 是抛物线 y=x^2<script type="math/tex">y=x^2</script> 上从点 O(0,0)<script type="math/tex">O(0,0)</script> 到点 A(2,4)<script type="math/tex">A(2,4)</script> 的一段弧.

解答: 由 \bex L:\ \left\{\ba{ll} x=x,\\ y=x^2, \ea\right. \quad x\mbox{ 从 } 0 \mbox{ 到 } 2, \eex<script type="math/tex; mode=display">\bex L:\ \left\{\ba{ll} x=x,\\ y=x^2, \ea\right. \quad x\mbox{ 从 } 0 \mbox{ 到 } 2, \eex</script> 我们有 \bex \int_L(x^2-y^2)\rd x =\int_0^2 (x^2-(x^2)^2)\rd x =-\cfrac{56}{15}. \eex<script type="math/tex; mode=display">\bex \int_L(x^2-y^2)\rd x =\int_0^2 (x^2-(x^2)^2)\rd x =-\cfrac{56}{15}. \eex</script>

3(2). 计算 \dps{\int_Ly\rd x}<script type="math/tex">\dps{\int_Ly\rd x}</script>, 其中 L<script type="math/tex">L</script> 是由直线 及 围成的矩形区域的整个边界 (按逆时针方向绕行).

解答: \bex \int_L y\rd x &=&\int_{(0,0)\to (4,0)}y\rd x +\int_{(4,0)\to (4,2)}y\rd x +\int_{(4,2)\to (0,2)}y\rd x +\int_{(0,2)\to (0,0)}y\rd x\\ &=&\int_{(4,2)\to (0,2)}y\rd x\\ &=&\int_4^0 2\rd x\quad\sex{\mbox{其余三项均为 }0}\\ &=&8. \eex<script type="math/tex; mode=display">\bex \int_L y\rd x &=&\int_{(0,0)\to (4,0)}y\rd x +\int_{(4,0)\to (4,2)}y\rd x +\int_{(4,2)\to (0,2)}y\rd x +\int_{(0,2)\to (0,0)}y\rd x\\ &=&\int_{(4,2)\to (0,2)}y\rd x\\ &=&\int_4^0 2\rd x\quad\sex{\mbox{其余三项均为 }0}\\ &=&8. \eex</script>

4. 计算 \dps{\int_L (x+y)\rd x+(y-x)\rd y}<script type="math/tex">\dps{\int_L (x+y)\rd x+(y-x)\rd y}</script>, 其中 L<script type="math/tex">L</script> 分别是 \item 抛物线 y^2=x<script type="math/tex">y^2=x</script> 上从点 (1,1)<script type="math/tex">(1,1)</script> 到点 (4,2)<script type="math/tex">(4,2)</script> 的一段弧; \item 从点 (1,1)<script type="math/tex">(1,1)</script> 到点 (4,2)<script type="math/tex">(4,2)</script> 的直线段.

解答: \item 由 \bex L:\ \left\{\ba{ll} x=y^2,\\ y=y, \ea\right. \quad x\mbox{ 从 } 1 \mbox{ 到 } 2, \eex<script type="math/tex; mode=display">\bex L:\ \left\{\ba{ll} x=y^2,\\ y=y, \ea\right. \quad x\mbox{ 从 } 1 \mbox{ 到 } 2, \eex</script> 我们有 \bex \int_L (x+y)\rd x+(y-x)\rd y &=&\int_1^2(y^2+y)\cdot2y\rd y +(y-y^2)\rd y\\ &=&\int_1^2(2y^3+y^2+y)\rd y\\ &=&\cfrac{34}{3}. \eex<script type="math/tex; mode=display">\bex \int_L (x+y)\rd x+(y-x)\rd y &=&\int_1^2(y^2+y)\cdot2y\rd y +(y-y^2)\rd y\\ &=&\int_1^2(2y^3+y^2+y)\rd y\\ &=&\cfrac{34}{3}. \eex</script> \item 由 \bex L:\ \left\{\ba{ll} x=x,\\ y=\cfrac{1}{3}x+\cfrac{2}{3}, \ea\right. \quad x\mbox{ 从 } 1 \mbox{ 到 } 4, \eex<script type="math/tex; mode=display">\bex L:\ \left\{\ba{ll} x=x,\\ y=\cfrac{1}{3}x+\cfrac{2}{3}, \ea\right. \quad x\mbox{ 从 } 1 \mbox{ 到 } 4, \eex</script> 我们有 \bex & &\int_L (x+y)\rd x+(y-x)\rd y\\ & &=\int_1^4 \sez{x+\sex{\cfrac{1}{3}x+\cfrac{2}{3}}}\rd x +\sez{\sex{\cfrac{1}{3}x+\cfrac{2}{3}}-x}\cdot\cfrac{1}{3}\rd x\\ & &=\int_1^4\sex{\cfrac{10}{9}x+\cfrac{8}{9}}\rd x\\ & &=11. \eex<script type="math/tex; mode=display">\bex & &\int_L (x+y)\rd x+(y-x)\rd y\\ & &=\int_1^4 \sez{x+\sex{\cfrac{1}{3}x+\cfrac{2}{3}}}\rd x +\sez{\sex{\cfrac{1}{3}x+\cfrac{2}{3}}-x}\cdot\cfrac{1}{3}\rd x\\ & &=\int_1^4\sex{\cfrac{10}{9}x+\cfrac{8}{9}}\rd x\\ & &=11. \eex</script>

9.6 格林公式及其应用

3(1). 利用 Green<script type="math/tex">Green</script> 公式计算 \dps{\oint_L(2x-y+4)\rd x+(3x+5y-6)\rd y}<script type="math/tex">\dps{\oint_L(2x-y+4)\rd x+(3x+5y-6)\rd y}</script>, 其中 L<script type="math/tex">L</script> 是以点 (0,0),(3,0),(3,2)<script type="math/tex">(0,0),(3,0),(3,2)</script> 为顶点的三角形区域 D<script type="math/tex">D</script> 的正向边界.

解答: 由 P=2x-y+4<script type="math/tex">P=2x-y+4</script>, Q=3x+5y-6<script type="math/tex">Q=3x+5y-6</script> 知 \bex \cfrac{\p P}{\p y}=-1,\quad \cfrac{\p Q}{\p x}=3. \eex<script type="math/tex; mode=display">\bex \cfrac{\p P}{\p y}=-1,\quad \cfrac{\p Q}{\p x}=3. \eex</script> 而 \bex \oint_LP\rd x+Q\rd y &=&\iint_D \sex{\cfrac{\p Q}{\p x}-\cfrac{\p P}{\p y}}\rd x\rd y\\ &=&4\iint_D \rd x\rd y\\ &=&4\cdot\sex{\cfrac{1}{2}\cdot 3\cdot 2}\\ &=&12. \eex<script type="math/tex; mode=display">\bex \oint_LP\rd x+Q\rd y &=&\iint_D \sex{\cfrac{\p Q}{\p x}-\cfrac{\p P}{\p y}}\rd x\rd y\\ &=&4\iint_D \rd x\rd y\\ &=&4\cdot\sex{\cfrac{1}{2}\cdot 3\cdot 2}\\ &=&12. \eex</script>

3(4). 利用 Green<script type="math/tex">Green</script> 公式计算 \dps{\oint_L(1-\cos y)\rd x-x(y-\sin y)\rd y}<script type="math/tex">\dps{\oint_L(1-\cos y)\rd x-x(y-\sin y)\rd y}</script>, 其中 L<script type="math/tex">L</script> 是正弦曲线 y=\sin x<script type="math/tex">y=\sin x</script> 上从点 (0,0)<script type="math/tex">(0,0)</script> 到点 (\pi,0)<script type="math/tex">(\pi,0)</script> 的一段弧.

解答: 设 L_1<script type="math/tex">L_1</script> 是从点 (0,0)<script type="math/tex">(0,0)</script> 到点 (\pi,0)<script type="math/tex">(\pi,0)</script> 的直线段, 与 L<script type="math/tex">L</script> 围成区域 D<script type="math/tex">D</script>. 再记 \bex P=1-\cos y,\quad Q=-x(y-\sin y), \eex<script type="math/tex; mode=display">\bex P=1-\cos y,\quad Q=-x(y-\sin y), \eex</script> 则 \bex \cfrac{\p P}{\p y}=\sin y,\quad \cfrac{\p Q}{\p x}=-y+\sin y. \eex<script type="math/tex; mode=display">\bex \cfrac{\p P}{\p y}=\sin y,\quad \cfrac{\p Q}{\p x}=-y+\sin y. \eex</script> 于是 \bex -\int_L+\int_{L_1}P\rd x+Q\rd y =\iint_D \sex{\cfrac{\p Q}{\p x}-\cfrac{\p P}{\p y}}\rd x\rd y, \eex<script type="math/tex; mode=display">\bex -\int_L+\int_{L_1}P\rd x+Q\rd y =\iint_D \sex{\cfrac{\p Q}{\p x}-\cfrac{\p P}{\p y}}\rd x\rd y, \eex</script> \bex \int_L P\rd x+Q\rd y &=&-\iint_D \sex{\cfrac{\p Q}{\p x}-\cfrac{\p P}{\p y}}\rd x\rd y +\int_{L_1}P\rd x+Q\rd y\\ &=&-\iint_D-y\rd x\rd y+\int_0^\pi (1-\cos 0)\rd x\\ &=&\int_0^\pi \rd x\int_0^{\sin x}y\rd y\\ &=&\cfrac{\pi}{4}. \eex<script type="math/tex; mode=display">\bex \int_L P\rd x+Q\rd y &=&-\iint_D \sex{\cfrac{\p Q}{\p x}-\cfrac{\p P}{\p y}}\rd x\rd y +\int_{L_1}P\rd x+Q\rd y\\ &=&-\iint_D-y\rd x\rd y+\int_0^\pi (1-\cos 0)\rd x\\ &=&\int_0^\pi \rd x\int_0^{\sin x}y\rd y\\ &=&\cfrac{\pi}{4}. \eex</script>

4(3). 证明曲线积分 \dps{\int_{(0,0)}^{(\pi,\pi)}(e^y+\sin x)\rd x+(xe^y-\cos y)\rd y}<script type="math/tex">\dps{\int_{(0,0)}^{(\pi,\pi)}(e^y+\sin x)\rd x+(xe^y-\cos y)\rd y}</script> 在 {\bf R}^2<script type="math/tex">{\bf R}^2</script> 内与路径无关, 并求其值.

解答: 设 P=e^y+\sin x<script type="math/tex">P=e^y+\sin x</script>, Q=xe^y-\cos y<script type="math/tex">Q=xe^y-\cos y</script>, 则 \bex \cfrac{\p P}{\p y}=e^y=\cfrac{\p Q}{\p x}, \eex<script type="math/tex; mode=display">\bex \cfrac{\p P}{\p y}=e^y=\cfrac{\p Q}{\p x}, \eex</script> 而 线积分 \dps{\int_{(0,0)}^{(\pi,\pi)}(e^y+\sin x)\rd x+(xe^y-\cos y)\rd y}<script type="math/tex">\dps{\int_{(0,0)}^{(\pi,\pi)}(e^y+\sin x)\rd x+(xe^y-\cos y)\rd y}</script> 在 {\bf R}^2<script type="math/tex">{\bf R}^2</script> 内与路径无关, 且 \bex \mbox{原曲线积分} &=&\int_{(0,0)\stackrel{y=x}{\to}(\pi,\pi)} (e^y+\sin x)\rd x+(xe^y-\cos y)\rd y\\ &=&\int_0^\pi (e^x+\sin x+xe^x-\cos x)\rd x\\ &=&\left.\sex{xe^x-\cos x-\sin x}\right|^\pi_0\\ &=&\pi e^\pi +2. \eex<script type="math/tex; mode=display">\bex \mbox{原曲线积分} &=&\int_{(0,0)\stackrel{y=x}{\to}(\pi,\pi)} (e^y+\sin x)\rd x+(xe^y-\cos y)\rd y\\ &=&\int_0^\pi (e^x+\sin x+xe^x-\cos x)\rd x\\ &=&\left.\sex{xe^x-\cos x-\sin x}\right|^\pi_0\\ &=&\pi e^\pi +2. \eex</script>

5(4). 验证曲线积分 \dps{(3x^2y^2+8xy^3)\rd x+(2x^3y+12x^2y^2+ye^y)\rd y}<script type="math/tex">\dps{(3x^2y^2+8xy^3)\rd x+(2x^3y+12x^2y^2+ye^y)\rd y}</script> 在 {\bf R}^2<script type="math/tex">{\bf R}^2</script> 内与是某个函数 u<script type="math/tex">u</script> 的全微分, 并求出这样的一个 u<script type="math/tex">u</script>.

解答: 设 P=3x^2y^2+8xy^3<script type="math/tex">P=3x^2y^2+8xy^3</script>, Q=2x^3y+12x^2y^2+ye^y<script type="math/tex">Q=2x^3y+12x^2y^2+ye^y</script>, 则 \bex \cfrac{\p P}{\p y}=6x^2y+24xy^2=\cfrac{\p Q}{\p y}, \eex<script type="math/tex; mode=display">\bex \cfrac{\p P}{\p y}=6x^2y+24xy^2=\cfrac{\p Q}{\p y}, \eex</script> 于是曲线积分 \dps{(3x^2y^2+8xy^3)\rd x+(2x^3y+12x^2y^2+ye^y)\rd y}<script type="math/tex">\dps{(3x^2y^2+8xy^3)\rd x+(2x^3y+12x^2y^2+ye^y)\rd y}</script> 在 {\bf R}^2<script type="math/tex">{\bf R}^2</script> 内与是某个函数 u<script type="math/tex">u</script> 的全微分, 且 \bex & &\cfrac{\p u}{\p x}=P=3x^2y^2+8xy^3\\ &\ra&u=x^3y^2+4x^2y^3+\varphi(y)\\ &\ra&\cfrac{\p u}{\p y}=2x^3y+12x^2y^2+\varphi‘(y)\\ &\ra&\varphi‘(y)=ye^y\\ &\ra&\varphi(y)=(y-1)e^y\\ &\ra&u=x^3y^2+4x^2y^3+(y-1)e^y+C. \eex<script type="math/tex; mode=display">\bex & &\cfrac{\p u}{\p x}=P=3x^2y^2+8xy^3\\ &\ra&u=x^3y^2+4x^2y^3+\varphi(y)\\ &\ra&\cfrac{\p u}{\p y}=2x^3y+12x^2y^2+\varphi‘(y)\\ &\ra&\varphi‘(y)=ye^y\\ &\ra&\varphi(y)=(y-1)e^y\\ &\ra&u=x^3y^2+4x^2y^3+(y-1)e^y+C. \eex</script>

10 无穷级数

10.1 常数项级数的概念与性质

3(2). 判断级数 \dps{\sum_{n=1}^\infty\ln\sex{1+\cfrac{1}{n}}}<script type="math/tex">\dps{\sum_{n=1}^\infty\ln\sex{1+\cfrac{1}{n}}}</script> 的收敛性.

解答: 由 \bex s_n&=&\sum_{k=1}^n\ln\sex{1+\cfrac{1}{k}} =\sum_{k=1}^n\ln\cfrac{k+1}{k} =\sum_{k=1}^n\sez{\ln(k+1)-\ln k}\\ &=&\ln (n+1)\to \infty\quad\sex{n\to\infty} \eex<script type="math/tex; mode=display">\bex s_n&=&\sum_{k=1}^n\ln\sex{1+\cfrac{1}{k}} =\sum_{k=1}^n\ln\cfrac{k+1}{k} =\sum_{k=1}^n\sez{\ln(k+1)-\ln k}\\ &=&\ln (n+1)\to \infty\quad\sex{n\to\infty} \eex</script> 知原级数收敛.

3(3). 判断级数 \dps{\sum_{n=1}^\infty\sqrt{\cfrac{n-1}{n+1}}}<script type="math/tex">\dps{\sum_{n=1}^\infty\sqrt{\cfrac{n-1}{n+1}}}</script> 的收敛性.

解答: 由 \bex \lim_{n\to\infty}\sqrt{\cfrac{n-1}{n+1}} =\sqrt{\lim_{n\to\infty}\cfrac{1-\cfrac{1}{n}}{1+\cfrac{1}{n}}} =1 \eex<script type="math/tex; mode=display">\bex \lim_{n\to\infty}\sqrt{\cfrac{n-1}{n+1}} =\sqrt{\lim_{n\to\infty}\cfrac{1-\cfrac{1}{n}}{1+\cfrac{1}{n}}} =1 \eex</script> 知原级数发散.

4(2). 若 \dps{\lim_{n\to\infty}a_n=a}<script type="math/tex">\dps{\lim_{n\to\infty}a_n=a}</script>, 求级数 \dps{\sum_{n=1}^\infty(a_n-a_{n+1})}<script type="math/tex">\dps{\sum_{n=1}^\infty(a_n-a_{n+1})}</script> 的和.

解答: \bex s_n=\sum_{k=1}^n(a_k-a_{k+1})=a_1-a_{n+1} \eex<script type="math/tex; mode=display">\bex s_n=\sum_{k=1}^n(a_k-a_{k+1})=a_1-a_{n+1} \eex</script> 故 \bex \sum_{n=1}^\infty(a_n-a_{n+1}) =\lim_{n\to\infty}s_n=a_1-a. \eex<script type="math/tex; mode=display">\bex \sum_{n=1}^\infty(a_n-a_{n+1}) =\lim_{n\to\infty}s_n=a_1-a. \eex</script>

5(4). 判断级数 \dps{\sum_{n=1}^\infty\sex{\cfrac{1}{4^n} -\cfrac{\ln^n3}{3^n}}}<script type="math/tex">\dps{\sum_{n=1}^\infty\sex{\cfrac{1}{4^n} -\cfrac{\ln^n3}{3^n}}}</script> 的收敛性.

解答: 由于 \bex \cfrac{1}{4}<1,\quad\cfrac{\ln3}{3}<1, \eex<script type="math/tex; mode=display">\bex \cfrac{1}{4}<1,\quad\cfrac{\ln3}{3}<1, \eex</script> 我们有 \bex \sum_{n=1}^\infty\cfrac{1}{4^n}\quad \sum_{n=1}^\infty\cfrac{\ln^n3}{3^n} \eex<script type="math/tex; mode=display">\bex \sum_{n=1}^\infty\cfrac{1}{4^n}\quad \sum_{n=1}^\infty\cfrac{\ln^n3}{3^n} \eex</script> 收敛, 而 \bex \sum_{n=1}^\infty\sex{\cfrac{1}{4^n} -\cfrac{\ln^n3}{3^n}} =\sum_{n=1}^\infty\cfrac{1}{4^n}-\sum_{n=1}^\infty\cfrac{\ln^n3}{3^n} \eex<script type="math/tex; mode=display">\bex \sum_{n=1}^\infty\sex{\cfrac{1}{4^n} -\cfrac{\ln^n3}{3^n}} =\sum_{n=1}^\infty\cfrac{1}{4^n}-\sum_{n=1}^\infty\cfrac{\ln^n3}{3^n} \eex</script> 收敛

10.2 常数项级数的审敛散法

1(8). \dps{\sum_{n=1}^\infty\sex{e^\cfrac{1}{\sqrt{n}}-1}}<script type="math/tex">\dps{\sum_{n=1}^\infty\sex{e^\cfrac{1}{\sqrt{n}}-1}}</script>.

解答: 由 \bex \lim_{n\to\infty}\cfrac{e^\cfrac{1}{\sqrt{n}}-1}{\cfrac{1}{\sqrt{n}}} =\lim_{x\to 0}\cfrac{e^x-1}{x} =\lim_{x\to 0}e^x=1 \eex<script type="math/tex; mode=display">\bex \lim_{n\to\infty}\cfrac{e^\cfrac{1}{\sqrt{n}}-1}{\cfrac{1}{\sqrt{n}}} =\lim_{x\to 0}\cfrac{e^x-1}{x} =\lim_{x\to 0}e^x=1 \eex</script> 及 \dps{\sum_{n=1}^\infty\cfrac{1}{\sqrt{n}}}<script type="math/tex">\dps{\sum_{n=1}^\infty\cfrac{1}{\sqrt{n}}}</script> 发散知原级数发散.

2(4). \dps{\sum_{n=1}^\infty n\tan \cfrac{\pi}{2^{n=1}}}<script type="math/tex">\dps{\sum_{n=1}^\infty n\tan \cfrac{\pi}{2^{n=1}}}</script>.

解答: 由 \bex \lim_{n\to\infty} \cfrac{(n+1)\tan \cfrac{\pi}{2^{n+1}}}{n\tan \cfrac{\pi}{2^{n=1}}} &=&\lim_{n\to\infty} \cfrac{n+1}{n}\cdot \cfrac{\cfrac{\pi}{2^{n+1}}}{\cfrac{\pi}{2^n}} =\cfrac{1}{2}<1 \eex<script type="math/tex; mode=display">\bex \lim_{n\to\infty} \cfrac{(n+1)\tan \cfrac{\pi}{2^{n+1}}}{n\tan \cfrac{\pi}{2^{n=1}}} &=&\lim_{n\to\infty} \cfrac{n+1}{n}\cdot \cfrac{\cfrac{\pi}{2^{n+1}}}{\cfrac{\pi}{2^n}} =\cfrac{1}{2}<1 \eex</script> 知原级数收敛.

3(6). \dps{\sum_{n=1}^\infty\cfrac{\sex{1+\cfrac{1}{n}}^{n^2}}{3^n}}<script type="math/tex">\dps{\sum_{n=1}^\infty\cfrac{\sex{1+\cfrac{1}{n}}^{n^2}}{3^n}}</script>.

解答: 由 \bex \lim_{n\to\infty}\sqrt[n]{\sex{1+\cfrac{1}{n}}^{n^2}{3^n}} =\lim_{n\to\infty}\cfrac{\sex{1+\cfrac{1}{n}}^n}{3} =\cfrac{e}{3}<1 \eex<script type="math/tex; mode=display">\bex \lim_{n\to\infty}\sqrt[n]{\sex{1+\cfrac{1}{n}}^{n^2}{3^n}} =\lim_{n\to\infty}\cfrac{\sex{1+\cfrac{1}{n}}^n}{3} =\cfrac{e}{3}<1 \eex</script> 知原级数收敛.

4(6). \dps{\sum_{n=1}^\infty \sin^n\sex{\cfrac{\pi}{4}+\cfrac{b}{n}}}<script type="math/tex">\dps{\sum_{n=1}^\infty \sin^n\sex{\cfrac{\pi}{4}+\cfrac{b}{n}}}</script>.

解答: 考察 \dps{\sum_{n=1}^\infty \sev{\sin^n\sex{\cfrac{\pi}{4}+\cfrac{b}{n}}}}<script type="math/tex">\dps{\sum_{n=1}^\infty \sev{\sin^n\sex{\cfrac{\pi}{4}+\cfrac{b}{n}}}}</script>, 由 \bex \lim_{n\to\infty}\sqrt[n]{\sev{\sin^n\sex{\cfrac{\pi}{4}+\cfrac{b}{n}}}} &=&\lim_{n\to\infty}\sev{\sin\sex{\cfrac{\pi}{4}+\cfrac{b}{n}}}\\ &=&\sin\sev{\lim_{n\to\infty}\sex{\cfrac{\pi}{4}+\cfrac{b}{n}}}\\ &=&\sin\cfrac{\pi}{4}=\cfrac{1}{\sqrt{2}}<1 \eex<script type="math/tex; mode=display">\bex \lim_{n\to\infty}\sqrt[n]{\sev{\sin^n\sex{\cfrac{\pi}{4}+\cfrac{b}{n}}}} &=&\lim_{n\to\infty}\sev{\sin\sex{\cfrac{\pi}{4}+\cfrac{b}{n}}}\\ &=&\sin\sev{\lim_{n\to\infty}\sex{\cfrac{\pi}{4}+\cfrac{b}{n}}}\\ &=&\sin\cfrac{\pi}{4}=\cfrac{1}{\sqrt{2}}<1 \eex</script> 知原级数绝对收敛.

4(7). \dps{\sum_{n=1}^\infty\cfrac{\ln(n!)}{n!}}<script type="math/tex">\dps{\sum_{n=1}^\infty\cfrac{\ln(n!)}{n!}}</script>.

解答: 由 \bex \cfrac{\ln(n!)}{n!} \leq \cfrac{n\ln n}{n!} =\cfrac{\ln n}{(n-1)!} \leq \cfrac{n-1}{(n-1)!} \leq \cfrac{1}{(n-2)!}\quad\sex{n\geq 2} \eex<script type="math/tex; mode=display">\bex \cfrac{\ln(n!)}{n!} \leq \cfrac{n\ln n}{n!} =\cfrac{\ln n}{(n-1)!} \leq \cfrac{n-1}{(n-1)!} \leq \cfrac{1}{(n-2)!}\quad\sex{n\geq 2} \eex</script> 知原级数收敛.

5(5). \dps{\sum_{n=1}^\infty(-1)^n\cfrac{n}{3^{n-1}}}<script type="math/tex">\dps{\sum_{n=1}^\infty(-1)^n\cfrac{n}{3^{n-1}}}</script>.

解答: 考察级数 \dps{\sum_{n=1}^\infty\cfrac{n}{3^{n-1}}}<script type="math/tex">\dps{\sum_{n=1}^\infty\cfrac{n}{3^{n-1}}}</script>, 由 \bex \lim_{n\to\infty}\cfrac{\cfrac{n+1}{3^n}}{\cfrac{n}{3^{n-1}}} =\cfrac{1}{3}<1 \eex<script type="math/tex; mode=display">\bex \lim_{n\to\infty}\cfrac{\cfrac{n+1}{3^n}}{\cfrac{n}{3^{n-1}}} =\cfrac{1}{3}<1 \eex</script> 知原级数绝对收敛.

5(7). \dps{\sum_{n=1}^\infty(-1)^n\cfrac{1}{\ln n}}<script type="math/tex">\dps{\sum_{n=1}^\infty(-1)^n\cfrac{1}{\ln n}}</script>.

解答: 由 Leibniz<script type="math/tex">Leibniz</script> 判别法知原级数收敛, 又由 \bex \lim_{n\to\infty}\cfrac{\cfrac{1}{\ln n}}{\cfrac{1}{n}} =\lim_{n\to\infty}\cfrac{1}{\cfrac{\ln n}{n}} =+\infty \eex<script type="math/tex; mode=display">\bex \lim_{n\to\infty}\cfrac{\cfrac{1}{\ln n}}{\cfrac{1}{n}} =\lim_{n\to\infty}\cfrac{1}{\cfrac{\ln n}{n}} =+\infty \eex</script> 知原级数条件收敛.

10.3 幂级数

1(7). 求幂级数 \dps{\sum_{n=1}^\infty (-1)^n\cfrac{x^n}{n^2}}<script type="math/tex">\dps{\sum_{n=1}^\infty (-1)^n\cfrac{x^n}{n^2}}</script> 的收敛半径与收敛域.

解答: 由 \dps{a_n=\cfrac{(-1)^n}{n^2}}<script type="math/tex">\dps{a_n=\cfrac{(-1)^n}{n^2}}</script> 知收敛半径 \bex R=\lim_{n\to\infty}\sev{\cfrac{a_n}{a_{n+1}}} =\lim_{n\to\infty}\cfrac{(n+1)^2}{n^2}=1. \eex<script type="math/tex; mode=display">\bex R=\lim_{n\to\infty}\sev{\cfrac{a_n}{a_{n+1}}} =\lim_{n\to\infty}\cfrac{(n+1)^2}{n^2}=1. \eex</script> 又当 x=-1<script type="math/tex">x=-1</script> 时, 原级数为 \dps{\sum_{n=1}^\infty\cfrac{1}{n^2}}<script type="math/tex">\dps{\sum_{n=1}^\infty\cfrac{1}{n^2}}</script>, 而收敛; 当 x=1<script type="math/tex">x=1</script> 时, 原级数为 \dps{\sum_{n=1}^\infty\cfrac{(-1)^n}{n^2}}<script type="math/tex">\dps{\sum_{n=1}^\infty\cfrac{(-1)^n}{n^2}}</script>, 而收敛. 故收敛域为 [-1,1]<script type="math/tex">[-1,1]</script>.

2(4). 求幂级数 \dps{\sum_{n=0}^\infty 2^n(x+a)^{2n}}<script type="math/tex">\dps{\sum_{n=0}^\infty 2^n(x+a)^{2n}}</script> 的收敛半径与收敛域, 其中 a<script type="math/tex">a</script> 为常数.

解答: 记 \dps{2(x+a)^2=t}<script type="math/tex">\dps{2(x+a)^2=t}</script>, 则由 \dps{\sum_{n=1}^\infty t^n}<script type="math/tex">\dps{\sum_{n=1}^\infty t^n}</script> 的收敛域为 (-1,1)<script type="math/tex">(-1,1)</script> 知 \bex -1<2(x+a)^2<1 \ra -a-\cfrac{1}{\sqrt{2}} <x<a+\cfrac{1}{\sqrt{2}}, \eex<script type="math/tex; mode=display">\bex -1<2(x+a)^2<1 \ra -a-\cfrac{1}{\sqrt{2}} \dps{\sex{-a-\cfrac{1}{\sqrt{2}},a+\cfrac{1}{\sqrt{2}}}}<script type="math/tex">\dps{\sex{-a-\cfrac{1}{\sqrt{2}},a+\cfrac{1}{\sqrt{2}}}}</script>, 收敛半径为 \dps{\cfrac{1}{\sqrt{2}}}<script type="math/tex">\dps{\cfrac{1}{\sqrt{2}}}</script>.

3(2). 求幂级数 \dps{\sum_{n=1}^\infty nx^{n-1}}<script type="math/tex">\dps{\sum_{n=1}^\infty nx^{n-1}}</script> 的和函数.

解答: \bex \sum_{n=1}^\infty nx^{n-1} &=&\sum_{n=1}^\infty (x^n)‘ =\sex{\sum_{n=1}^\infty x^n}‘\\ &=&\sex{\cfrac{x}{1-x}}‘ =\sex{-1+\cfrac{1}{1-x}}‘\\ &=&\cfrac{1}{(x-1)^2},\quad x\in (-1,1). \eex<script type="math/tex; mode=display">\bex \sum_{n=1}^\infty nx^{n-1} &=&\sum_{n=1}^\infty (x^n)‘ =\sex{\sum_{n=1}^\infty x^n}‘\\ &=&\sex{\cfrac{x}{1-x}}‘ =\sex{-1+\cfrac{1}{1-x}}‘\\ &=&\cfrac{1}{(x-1)^2},\quad x\in (-1,1). \eex</script>

10.4 函数展开成幂级数

2(4). 将函数 \dps{\cfrac{1}{x^2-5x+6}}<script type="math/tex">\dps{\cfrac{1}{x^2-5x+6}}</script> 展开成 x<script type="math/tex">x</script> 的幂级数, 并指出展开式成立的区间.

解答: \bex \cfrac{1}{x^2-5x+6} &=&\cfrac{1}{(x-2)(x-3)}\\ &=&\cfrac{1}{x-3}-\cfrac{1}{x-2}\\ &=&-\cfrac{1}{3}\cdot \cfrac{1}{1-\cfrac{x}{3}} +\cfrac{1}{2}\cdot \cfrac{1}{1-\cfrac{x}{2}}\\ &=&-\cfrac{1}{3} \sum_{n=0}^\infty \sex{\cfrac{x}{3}}^n +\cfrac{1}{2} \sum_{n=0}^\infty \sex{\cfrac{x}{2}}^n\\ & &\sex{\left.\ba{ll} \sev{\cfrac{x}{3}}<1\\ \sev{\cfrac{x}{2}}<1 \ea\right\} \ra \sev{x}<2 }\\ &=&\sum_{n=0}^\infty \sex{-\cfrac{1}{3^{n+1}} +\cfrac{1}{2^{n+1}}} x^n\quad\sex{-2<x<2}. \eex<script type="math/tex; mode=display">\bex \cfrac{1}{x^2-5x+6} &=&\cfrac{1}{(x-2)(x-3)}\\ &=&\cfrac{1}{x-3}-\cfrac{1}{x-2}\\ &=&-\cfrac{1}{3}\cdot \cfrac{1}{1-\cfrac{x}{3}} +\cfrac{1}{2}\cdot \cfrac{1}{1-\cfrac{x}{2}}\\ &=&-\cfrac{1}{3} \sum_{n=0}^\infty \sex{\cfrac{x}{3}}^n +\cfrac{1}{2} \sum_{n=0}^\infty \sex{\cfrac{x}{2}}^n\\ & &\sex{\left.\ba{ll} \sev{\cfrac{x}{3}}<1\\ \sev{\cfrac{x}{2}}<1 \ea\right\} \ra \sev{x}<2 }\\ &=&\sum_{n=0}^\infty \sex{-\cfrac{1}{3^{n+1}} +\cfrac{1}{2^{n+1}}} x^n\quad\sex{-2

2(7). 将函数 \dps{\cfrac{x}{4+x^2}}<script type="math/tex">\dps{\cfrac{x}{4+x^2}}</script> 展开成 x<script type="math/tex">x</script> 的幂级数, 并指出展开式成立的区间.

解答: \bex \cfrac{x}{4+x^2}&=& \cfrac{x}{4}\cdot \cfrac{1}{1-\sex{-\cfrac{x^2}{4}}}\\ &=&\cfrac{x}{4}\sum_{n=0}^\infty \sex{-\cfrac{x^2}{4}}^n\quad \sex{\sev{-\cfrac{x^2}{4}}<1\ra \sev{x}<2}\\ &=&\sum_{n=0}^\infty \cfrac{(-1)^n}{4^{n+1}}x^{2n+1}\quad \sex{-2<x<2}. \eex<script type="math/tex; mode=display">\bex \cfrac{x}{4+x^2}&=& \cfrac{x}{4}\cdot \cfrac{1}{1-\sex{-\cfrac{x^2}{4}}}\\ &=&\cfrac{x}{4}\sum_{n=0}^\infty \sex{-\cfrac{x^2}{4}}^n\quad \sex{\sev{-\cfrac{x^2}{4}}<1\ra \sev{x}<2}\\ &=&\sum_{n=0}^\infty \cfrac{(-1)^n}{4^{n+1}}x^{2n+1}\quad \sex{-2

4. 将函数 f(x)=e^x<script type="math/tex">f(x)=e^x</script> 展开为 x-1<script type="math/tex">x-1</script> 的幂级数.

解答: 由于 f(x)=e^x<script type="math/tex">f(x)=e^x</script>, f^{(n)}(x)=e^x\ (n\geq 1)<script type="math/tex">f^{(n)}(x)=e^x\ (n\geq 1)</script>, 我们有 \bex f^{(n)}(1)=e, \eex<script type="math/tex; mode=display">\bex f^{(n)}(1)=e, \eex</script> 而 f(x)<script type="math/tex">f(x)</script> 在 x=1<script type="math/tex">x=1</script> 处的 Taylor<script type="math/tex">Taylor</script> 级数为 \bex \sum_{n=0}^\infty \cfrac{e}{n!}(x-1)^n =e\sum_{n=0}^\infty \cfrac{(x-1)^n}{n!}, \eex<script type="math/tex; mode=display">\bex \sum_{n=0}^\infty \cfrac{e}{n!}(x-1)^n =e\sum_{n=0}^\infty \cfrac{(x-1)^n}{n!}, \eex</script> 其收敛半径 R=\infty<script type="math/tex">R=\infty</script>. 又 \bex \sev{R_n(x)} &=&\sev{\cfrac{f^{(n+1)}(\xi)}{(n+1)!}(x-1)^{n+1}}\quad\sex{\xi\mbox{ 在 } 1,\ x\mbox{ 之间}}\\ &=&\cfrac{e^\xi}{(n+1)!}\sev{x-1}^{n+1}\\ &\leq&e^{\max\sed{1,\sev{x}}}\cdot \cfrac{\sev{x-1}^{n+1}}{(n+1)!}\\ &\to& 0\quad\sex{n\to\infty}, \eex<script type="math/tex; mode=display">\bex \sev{R_n(x)} &=&\sev{\cfrac{f^{(n+1)}(\xi)}{(n+1)!}(x-1)^{n+1}}\quad\sex{\xi\mbox{ 在 } 1,\ x\mbox{ 之间}}\\ &=&\cfrac{e^\xi}{(n+1)!}\sev{x-1}^{n+1}\\ &\leq&e^{\max\sed{1,\sev{x}}}\cdot \cfrac{\sev{x-1}^{n+1}}{(n+1)!}\\ &\to& 0\quad\sex{n\to\infty}, \eex</script> 我们得到 \bex f(x)=e\sum_{n=0}^\infty \cfrac{(x-1)^n}{n!},\quad -\infty<x<\infty. \eex<script type="math/tex; mode=display">\bex f(x)=e\sum_{n=0}^\infty \cfrac{(x-1)^n}{n!},\quad -\infty

10.5 傅里叶级数