張宇1000題高等數學 第十八章 多元函數積分學（二）

$C$組

1.設$y'=f(x,y)$是一條簡單封閉曲線$L$（取正向），$f(x,y)\ne0$，其所圍成區域記爲$D$，$D$的面積爲$1$，則$I=\displaystyle\oint_Lxf(x,y)\mathrm{d}x-\cfrac{y}{f(x,y)}\mathrm{d}y=$______。

解  將邊界方程代入被積函數，於是有
$\begin{aligned} I&=\displaystyle\oint_Lxy'\mathrm{d}x-\cfrac{y}{y'}\mathrm{d}y=\displaystyle\oint_Lx\mathrm{d}y-y\mathrm{d}x\\ &=\displaystyle\iint\limits_{D}(1+1)\mathrm{d}\sigma=2. \end{aligned}$

2.設$f(x)$在$[0,1]$上連續，證明：$\displaystyle\int^1_0\mathrm{d}x\displaystyle\int^x_0\mathrm{d}y\displaystyle\int^y_0f(x)f(y)f(z)\mathrm{d}z=\cfrac{1}{3!}\left[\displaystyle\int^1_0f(t)\mathrm{d}t\right]$。

解  因爲$f(x)$在$[0,1]$上連續，所以在$[0,1]$上存在原函數。
  設$F'(t)=f(t)(t\in[0,1])$，則
$\begin{aligned} \displaystyle\int^1_0\mathrm{d}x\displaystyle\int^x_0\mathrm{d}y\displaystyle\int^y_0f(x)f(y)f(z)\mathrm{d}z&=\displaystyle\int^1_0f(x)\mathrm{d}x\displaystyle\int^x_0f(y)[F(y)-F(0)]\mathrm{d}y\\ &=\displaystyle\int^1_0f(x)\mathrm{d}x\displaystyle\int^x_0[F(y)-F(0)]\mathrm{d}[F(y)-F(0)]\\ &=\displaystyle\int^1_0f(x)\cfrac{[F(y)-F(0)]^2}{2}\biggm\vert^x_0\mathrm{d}x\\ &=\cfrac{1}{2}\displaystyle\int^1_0f(x)[F(x)-F(0)]^2\mathrm{d}x\\ &=\cfrac{1}{2}\displaystyle\int^1_0f(x)[F(x)-F(0)]^2\mathrm{d}[F(x)-F(0)]\\ &=\cfrac{1}{3!}[F(x)-F(0)]^3\biggm\vert^1_0=\cfrac{1}{3!}[F(1)-F(0)]^3\\ &=\cfrac{1}{3!}\left[\displaystyle\int^1_0f(t)\mathrm{d}t\right]. \end{aligned}$

11.設$f(x,y)$在全平面有連續偏導數，曲線積分$\displaystyle\int_Lf(x,y)\mathrm{d}x+x\cos y\mathrm{d}y$在全平面與路徑無關，且$\displaystyle\int^{(t,t^2)}_{(0,0)}f(x,y)\mathrm{d}x+x\cos y\mathrm{d}y=t^2$，求$f(x,y)$。

解  $\displaystyle\int_Lf(x,y)\mathrm{d}x+x\cos y\mathrm{d}y$在全平面與路徑無關$\Leftrightarrow\cfrac{\partial}{\partial x}(x\cos y)=\cfrac{\partial f}{\partial y}$，即$\cfrac{\partial f}{\partial y}=\cos y$，積分得$f(x,y)=\sin y+C$。
  因
$\begin{aligned} f(x,y)\mathrm{d}x+x\cos y\mathrm{d}y&=\sin y\mathrm{d}x+C(x)\mathrm{d}x+x\cos y\mathrm{d}y\\ &=\sin y\mathrm{d}x+x\mathrm{d}\sin y+\mathrm{d}\left[\displaystyle\int^x_0C(s)\mathrm{d}s\right]=\mathrm{d}\left[x\sin y+\displaystyle\int^x_0C(s)\mathrm{d}s\right], \end{aligned}$
  則有$\left[x\sin y+\displaystyle\int^x_0C(s)\mathrm{d}s\right]\biggm\vert^{(t,t^2)}_{(0,0)}=t^2$，即$t\sin t^2+\displaystyle\int^x_0C(s)\mathrm{d}s=t^2\Rightarrow\sin t^2+2t^2\cos t^2+C(t)=2t$，因此$f(x,y)=\sin y+2x-\sin x^2-2x^2\cos x^2$。

14.設$f(x),g(x)$二階導數連續，且$f(0)=-2,g(0)=0$。對於任意一條逐段光滑的封閉曲線$L$，有$\displaystyle\oint_L2[xf(x)+g(y)]\mathrm{d}x+[x^2g(y)+2xy^2+2xf(y)]\mathrm{d}y=0$。

（1）求$f(x),g(x)$；

解  由曲線積分與路徑無關，有$\cfrac{\partial P}{\partial y}=\cfrac{\partial Q}{\partial x}$，即$2xg(y)+2y^2+2f(y)=2xf'(y)+2g'(y)$，也即$2x[g(y)-f'(y)]=-2y^2-2f(y)+2g'(y)$，比較等式兩邊$x$的同冪次係數，有
$\begin{cases} g(y)-f'(y)=0,&\qquad(1)\\ -2y^2-2f(y)+2g'(y)=0,&\qquad(2) \end{cases}$
  將$(1)$兩邊同時對$y$求導，得$g'(y)=f''(y)$，代入$(2)$，有$f''(y)-f(y)=y^2$。
  其特徵方程爲$r^2-1=0$，得$r_1=1,r_2=-1$，於是齊次方程的通解爲$f_1(y)=C_1e^y+C_2e^{-y}(3)$。
  設特解爲$f^*(y)=ay^2+by+c$，代入$(3)$，有$a=-1,b=0,c=-2$，故$f^*(y)=-y^2-2$，於是通解爲$f(y)=C_1e^y+C_2e^{-y}-y^2-2$，由$f(0)=-2$，有$C_1+C_2=0$，又$g(y)=f'(y)=C_1e^y-C_2e^{-y}-2y$，由$g(0)=0$，有$C_1-C_2=0$，故$C_1=C_2=0$。所以$f(y)=-y^2-2,g(y)=-2y$，也即$f(x)=-x^2-2,g(x)=-2x$。

（2）計算$\displaystyle\int^{(0,0)}_{(1,1)}2[xf(x)+g(y)]\mathrm{d}x+[x^2g(y)+2xy^2+2xf(y)]\mathrm{d}y$。

解  依題設，結合（1），由折線法，$(1,1)\to(0,1)\to(0,0)$，得
$\begin{aligned} &\displaystyle\int^{(0,0)}_{(1,1)}2[xf(x)+g(y)]\mathrm{d}x+[x^2g(y)+2xy^2+2xf(y)]\mathrm{d}y\\ =&\displaystyle\int^{(0,0)}_{(1,1)}2[x(-y^2-2)-2y]\mathrm{d}x+[x^2(-2y)+2xy^2+2x(-y^2-2)]\mathrm{d}y\\ =&2\displaystyle\int^1_0(3x+2)\mathrm{d}x=7. \end{aligned}$

15.設$f(x,y,z)$爲連續函數，$\Sigma$爲曲面$z=\cfrac{1}{2}(x^2+y^2)$介於$z=2$與$z=8$之間的上側部分，求$\displaystyle\iint\limits_{\Sigma}[yf(x,y,z)+x]\mathrm{d}y\mathrm{d}z+[xf(x,y,z)+y]\mathrm{d}x\mathrm{d}z+[2xyf(x,y,z)+z]\mathrm{d}y\mathrm{d}x$。

解  $\Sigma$的單位法向量爲$\bm{n}^\circ=\left(-\cfrac{x}{\sqrt{x^2+y^2+1}},-\cfrac{y}{\sqrt{x^2+y^2+1}},\cfrac{1}{\sqrt{x^2+y^2+1}}\right)$，將原給第二型曲面積分化成第一型曲面積分，得原積分$=\displaystyle\iint\limits_{\Sigma}\cfrac{-x^2-y^2-z}{\sqrt{x^2+y^2+z^2}}\mathrm{d}S$。
  再將$\Sigma$投影到平面$xOy$，得投影域$D=\{(x,y)|2\leqslant\sqrt{x^2+y^2}\leqslant4\}$。又$\mathrm{d}S=\sqrt{\left(\cfrac{\partial z}{\partial x}\right)^2+\left(\cfrac{\partial z}{\partial y}\right)^2+1}\mathrm{d}\sigma=\sqrt{x^2+y^2+1}\mathrm{d}\sigma$，從而原積分$=-\cfrac{1}{2}\displaystyle\iint\limits_{D}(x^2+y^2)\mathrm{d}\sigma=-\cfrac{1}{2}\displaystyle\int^{2\pi}_0\mathrm{d}\theta\displaystyle\int^4_2r^3\mathrm{d}r=-60\pi$。

19.設函數$f(x,y,z)$在區域$\Omega=\{(x,y,z)|x^2+y^2+z^2\leqslant1\}$上具有二階偏導數，且滿足$\cfrac{\partial^2f}{\partial x^2}+\cfrac{\partial^2f}{\partial y^2}+\cfrac{\partial^2f}{\partial z^2}=\sqrt{x^2+y^2+z^2}$，計算$I=\displaystyle\iiint\limits_{\Omega}\left(x\cfrac{\partial f}{\partial x}+y\cfrac{\partial f}{\partial y}+z\cfrac{\partial f}{\partial z}\right)\mathrm{d}x\mathrm{d}y\mathrm{d}z$。

解  設球面$\Sigma:x^2+y^2+z^2=1$外側的方向餘弦爲$(\cos\alpha,\cos\beta,\cos\gamma)$，有$\cfrac{\mathrm{d}y\mathrm{d}z}{\cos\alpha}=\cfrac{\mathrm{d}x\mathrm{d}z}{\cos\beta}=\cfrac{\mathrm{d}y\mathrm{d}x}{\cos\gamma}=\mathrm{d}S$，故
$\widetilde{I}=\displaystyle\oiint\limits_{\Sigma}\left(\cfrac{\partial f}{\partial x}\cos\alpha+\cfrac{\partial f}{\partial y}\cos\beta+\cfrac{\partial f}{\partial z}\cos\gamma\right)\mathrm{d}S=\displaystyle\oiint\limits_{\Sigma}\cfrac{\partial f}{\partial x}\mathrm{d}y\mathrm{d}z+\cfrac{\partial f}{\partial y}\mathrm{d}z\mathrm{d}x+\cfrac{\partial f}{\partial z}\mathrm{d}x\mathrm{d}y.\qquad(1)$
  又由於$\Sigma:x^2+y^2+z^2=1$，故還可寫出
$\begin{aligned} \widetilde{I}&=\displaystyle\oiint\limits_{\Sigma}(x^2+y^2+z^2)\left(\cfrac{\partial f}{\partial x}\cos\alpha+\cfrac{\partial f}{\partial y}\cos\beta+\cfrac{\partial f}{\partial z}\cos\gamma\right)\mathrm{d}S\\ &=\displaystyle\oiint\limits_{\Sigma}(x^2+y^2+z^2)\cfrac{\partial f}{\partial x}\mathrm{d}y\mathrm{d}z+(x^2+y^2+z^2)\cfrac{\partial f}{\partial y}\mathrm{d}z\mathrm{d}x+(x^2+y^2+z^2)\cfrac{\partial f}{\partial z}\mathrm{d}x\mathrm{d}y.\qquad(2) \end{aligned}$
  $(1),(2)$式都用高斯公式，有
$\begin{aligned} &\displaystyle\iiint\limits_{\Omega}\left(\cfrac{\partial^2f}{\partial x^2}+\cfrac{\partial^2f}{\partial y^2}+\cfrac{\partial^2f}{\partial z^2}\right)\mathrm{d}v\\ =&2\displaystyle\iiint\limits_{\Omega}\left(x\cfrac{\partial f}{\partial x}+y\cfrac{\partial f}{\partial y}+z\cfrac{\partial f}{\partial z}\right)\mathrm{d}x\mathrm{d}y\mathrm{d}z+\displaystyle\iiint\limits_{\Omega}(x^2+y^2+z^2)\left(\cfrac{\partial^2f}{\partial x^2}+\cfrac{\partial^2f}{\partial y^2}+\cfrac{\partial^2f}{\partial z^2}\right)\mathrm{d}v \end{aligned}$
  故
$\begin{aligned} I&=\cfrac{1}{2}\displaystyle\iiint\limits_{\Omega}\left(\cfrac{\partial^2f}{\partial x^2}+\cfrac{\partial^2f}{\partial y^2}+\cfrac{\partial^2f}{\partial z^2}\right)\mathrm{d}v-\cfrac{1}{2}\displaystyle\iiint\limits_{\Omega}(x^2+y^2+z^2)\left(\cfrac{\partial^2f}{\partial x^2}+\cfrac{\partial^2f}{\partial y^2}+\cfrac{\partial^2f}{\partial z^2}\right)\mathrm{d}v\\ &=\cfrac{1}{2}\displaystyle\iiint\limits_{\Omega}\sqrt{x^2+y^2+z^2}\mathrm{d}v-\cfrac{1}{2}\displaystyle\iiint\limits_{\Omega}(x^2+y^2+z^2)^{\frac{3}{2}}\mathrm{d}v\\ &=\cfrac{1}{2}\displaystyle\int^{2\pi}_0\mathrm{d}\theta\displaystyle\int^\pi_0\mathrm{d}\varphi\displaystyle\int^1_0r^3\sin\varphi\mathrm{d}r-\cfrac{1}{2}\displaystyle\int^{2\pi}_0\mathrm{d}\theta\displaystyle\int^\pi_0\mathrm{d}\varphi\displaystyle\int^1_0r^5\sin\varphi\mathrm{d}r\\ &=2\pi\left(\cfrac{1}{4}-\cfrac{1}{6}\right). \end{aligned}$

20.計算曲線積分$I=\displaystyle\oint_Ly^2\mathrm{d}x+z^2\mathrm{d}y+x^2\mathrm{d}z$，其中曲線$L$爲$\begin{cases}x^2+y^2+z^2=4,\\x^2+y^2=2x\end{cases}(z\geqslant0)$，從$x$軸的正向往負向看去，取逆時針方向。

解  $\Sigma:z=\sqrt{4-x^2-y^2}$，以$L$爲邊界，取上側，由斯托克斯公式，
$\begin{aligned} I&=\displaystyle\oint_Ly^2\mathrm{d}x+z^2\mathrm{d}y+x^2\mathrm{d}z\\& =\displaystyle\iint\limits_{\Sigma}\begin{vmatrix}\mathrm{d}y\mathrm{d}z&\mathrm{d}z\mathrm{d}x&\mathrm{d}x\mathrm{d}y\\\cfrac{\partial}{\partial x}&\cfrac{\partial}{\partial y}&\cfrac{\partial}{\partial z}\\y^2&z^2&x^2\end{vmatrix}\\ &=\displaystyle\iint\limits_{\Sigma}-2z\mathrm{d}y\mathrm{d}z-2x\mathrm{d}z\mathrm{d}x-2y\mathrm{d}x\mathrm{d}y \end{aligned}$
  $\Sigma$的法向量爲
$\bm{n}^\circ=(-z'_x,-z'_y,1)=\left(\cfrac{x}{\sqrt{4-x^2-y^2}},\cfrac{y}{\sqrt{4-x^2-y^2}},1\right),\\ D_{xy}=\{(x,y)|(x-1)^2+y^2\leqslant1\},$
  故
$\begin{aligned} I&=\displaystyle\iint\limits_{D_{xy}}(-2\sqrt{4-x^2-y^2},-2x,-2y)\cdot\left(\cfrac{x}{\sqrt{4-x^2-y^2}},\cfrac{y}{\sqrt{4-x^2-y^2}},1\right)\mathrm{d}x\mathrm{d}y\\ &=\displaystyle\iint\limits_{D_{xy}}\left(-2x-\cfrac{2xy}{\sqrt{4-x^2-y^2}}-2y\right)\mathrm{d}x\mathrm{d}y=-2\displaystyle\iint\limits_{D_{xy}}x\mathrm{d}x\mathrm{d}y\\ &=-2\displaystyle\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\mathrm{d}\theta\displaystyle\int^{2\cos\theta}r\cos\theta\cdot r\mathrm{d}r=-\cfrac{32}{3}\displaystyle\int^{\frac{\pi}{2}}_0\cos^4\theta\mathrm{d}\theta=-2\pi. \end{aligned}$

寫在最後

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