曲面and曲线积分3

$ 1.设连续可微函数z=f(x,y)由方程F(xz-y,x-yz)=0（其中F(u,v)有连续的偏导数）唯一确定，L为正向单位圆周。试求I=\oint_{L}(xz^2+2yz)dy-(2xz+yz^2)dx。$

• \(第二型曲线积分的做法就那么几种，要么参数化，要么试试格林公式或者斯托克斯公式。\)
• \(这里显然没法参数化，那就试试格林公式。D:x^2+y^2 \leq 1\)

\[I=\iint_D (\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dxdy=\iint_D[(z^2+2xz\frac{\partial z}{\partial x}+2y\frac{\partial z}{\partial x})-(-2x \frac{\partial z}{\partial y}-z^2-2zy\frac{\partial z}{\partial y})]dxdy \]

• \(\frac{\partial z}{\partial x}和\frac{\partial z}{\partial y}怎么得到？对F做全微分：\)

\[0=(F_1'z+F_2')dx+(-F_1'-F_2'z)dy+(F_1'x-F_2'y)dz \]

• \(所以\)

\[\frac{\partial z}{\partial x}=\frac{F_1'z+F_2'}{F_2'y-F_1'x} \]

\[\frac{\partial z}{\partial y}=\frac{-F_1'-F_2'z}{F_2'y-F_1'x} \]

• \(代入上面的式子\)

\[I=\iint_D 2z^2+\frac{2xz^2F_1'+2xzF_2'+2yzF_1'+2yF_2'-(2xF_1'+2xzF_2'+2zyF_1'+2z^2yF_2')}{F_2'y-F_1'x}dxdy \]

\[I=2\iint_D dxdy=2\pi \]

\(2.设L是圆x^2+y^2=a^2的正向，求I=\oint_L \frac{-ydx+xdy}{Ax^2+2Bxy+Cy^2}(A>0,AC-B^2>0)。\)

• \(猜测\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}，验证一下：\)

\[\frac{\partial Q}{\partial x}=\frac{(Ax^2+2Bxy+Cy^2)-x(2Ax+2By)}{(Ax^2+2Bxy+Cy^2)^2}=\frac{Cy^2-Ax^2}{(Ax^2+2Bxy+Cy^2)^2} \]

\[\frac{\partial P}{\partial y}=\frac{-(Ax^2+2Bxy+Cy^2)+y(2Cy+2Bx)}{(Ax^2+2Bxy+Cy^2)^2}=\frac{Cy^2-Ax^2}{(Ax^2+2Bxy+Cy^2)^2} \]

• \(套格林公式。注意(0,0)是奇点，作辅助曲线L_1:Ax^2+2Bxy+Cy^2=r^2（r>0），逆时针方向。\)

\[I=\oint_{L+L_{1}^{-}}+\oint_{L_1}=\iint_{D}0dxdy+\oint_{L_1}=\oint_{L_1} \]

\[I=\oint_{L_1}=\frac{-ydx+xdy}{r^2}=\frac{1}{r^2}\oint_{L_1} -ydx+xdy \]

• \(再用一次格林公式\)

\[I=\frac{1}{r^2}\oint_{L_1} -ydx+xdy=\frac{1}{r^2}\iint_{Ax^2+2Bxy+Cy^2 \leq r^2}2dxdy \]

• \(于是问题转化为算Ax^2+2Bxy+Cy^2 \leq r^2的面积。\)

\[Ax^2+2Bxy+Cy^2=(\sqrt{A}x+\frac{B}{\sqrt{A}}y)^2+(\sqrt{C-\frac{B^2}{A}}y)^2 \]

• \(作换元\)

\[\left \{ \begin{array}{rcl} &u=\sqrt{A}x+\frac{B}{\sqrt{A}}y \\ &v=\sqrt{C-\frac{B^2}{A}}y \end{array} \right.\]

\[\frac{\partial (x,y)}{\partial (u,v)}=\frac{1}{\frac{\partial (u,v)}{\partial (x,y)}}=\frac{1}{ \begin{vmatrix} \sqrt{A} & 0 \\ \frac{B}{\sqrt{A}} & \sqrt{C-\frac{B^2}{A}} \end{vmatrix} }=\frac{1}{\sqrt{AC-B^2}}\]

\[S=\frac{1}{\sqrt{AC-B^2}} \cdot \pi r^2 \]

• \(所以\)

\[I=\frac{1}{r^2} \cdot 2S=\frac{2\pi}{\sqrt{AC-B^2}} \]

\(3.设I_a(r)=\oint_C \frac{ydx-xdy}{(x^2+y^2)^a}，其中a为常数，曲线C为椭圆x^2+xy+y^2=r^2,取正向。求极限\lim_{r \to \infty}I_a(r)。\)

• \(依旧是猜测用格林公式：\)

\[\frac{\partial Q}{\partial x}=\frac{-(x^2+y^2)^a+2ax^2(x^2+y^2)^{a-1}}{(x^2+y^2)^{2a}}=\frac{2ax^2-x^2-y^2}{(x^2+y^2)^{a+1}} \]

\[\frac{\partial P}{\partial y}=\frac{(x^2+y^2)^a-2ay^2(x^2+y^2)^{a-1}}{(x^2+y^2)^{2a}}=\frac{x^2+y^2-2ay^2}{(x^2+y^2)^{a+1}} \]

\[\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=\frac{(2a-2)}{(x^2+y^2)^{a}} \]

\[I_a(r)=\iint_{x^2+xy+y^2 \leq r^2}\frac{(2a-2)}{(x^2+y^2)^{a}}dxdy \]

• \(尝试换元\)

\[x^2+xy+y^2=(x+\frac{1}{2}y)^2+(\frac{\sqrt{3}}{2}y)^2 \]

\[\left \{ \begin{array}{rcl} & u=x+\frac{1}{2}y \\ & v=\frac{\sqrt{3}}{2}y \end{array} \right. \]

\[\frac{\partial (x,y)}{\partial (u,v)}=\frac{1}{\frac{\partial (u,v)}{\partial (x,y)}}=\frac{1}{\begin{vmatrix}1 & 0 \\ \frac{1}{2} & \frac{\sqrt{3}}{2}\end{vmatrix}}=\frac{2}{\sqrt{3}} \]

\[I_a(r)=\frac{2}{\sqrt{3}}(2a-2)\iint_{u^2+v^2 \leq r^2}\frac{1}{((u-\frac{1}{\sqrt{3}}v)^2+(\frac{2}{\sqrt{3}}v)^2)^a}dudv \]

\[I_a(r)=\frac{2}{\sqrt{3}}(2a-2)\iint_{u^2+v^2 \leq r^2}\frac{1}{(u^2+\frac{5}{3}v^2-\frac{2}{\sqrt{3}}uv)^a}dudv \]

• \(再换元\)

\[\left \{ \begin{array}{rcl} & u=kcos\theta \\ & v=ksin\theta \end{array} \right. \]

\[\frac{\partial (u,v)}{\partial (k,\theta)}=k \]

\[I_a(r)=\frac{2}{\sqrt{3}}(2a-2)\int_{0}^{r}dk\int_{0}^{2\pi}\frac{k}{k^{2a}(cos^2\theta+\frac{5}{3}sin^2\theta-\frac{2}{\sqrt{3}}cos\theta sin\theta)^a}d\theta \]

\[I_a(r)=\frac{2}{\sqrt{3}}r^{2-2a}\int_{0}^{2\pi}\frac{1}{(cos^2\theta+\frac{5}{3}sin^2\theta-\frac{2}{\sqrt{3}}cos\theta sin\theta)^a}d\theta \]

• \(分情况考虑\)

\[a<1,\lim_{r \to +\infty}I_a(r) \to +\infty \]

\[a>1,\lim_{r \to +\infty}I_a(r) \to 0 \]

\[a=1,I_1(r)=\oint_C \frac{ydx-xdy}{(x^2+y^2)}=-2\pi \]