VC4.24

From Exampleproblems

Project the given sphere S on the xy-plane in the region R bounded by the circle $x^2+y^2=1,z=0\,$

Hence $\mathrm{curl}F=\begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y & z & x \end{vmatrix}\,$

= $[\frac{\partial x}{\partial y}-\frac{\partial z}{\partial z}]i-[\frac{\partial x}{\partial x}-\frac{\partial y}{\partial z}]j+[\frac{\partial z}{\partial x}-\frac{\partial y}{\partial y}]k=-(i+j+k)\,$ --(1)

Given surface S is $\phi(x,y,z)=x^2+y^2+z^2=1\,$ --(2)

From(2), $\nabla\phi=[i\frac{\partial}{\partial x}+j\frac{\partial}{\partial y}+k\frac{\partial}{\partial z}](x^2+y^2+z^2)=2(xi+yj+zk)\,$

Therefore,n=unit normal vector to surface(2)= $\frac{\nabla\phi}{|\nabla\phi|}=\frac{2(xi+yj+zk)}{\sqrt{(4x^2+4y^2+4z^2)}}=xi+yj+zk\,$ from (2)

Now, $n\cdot k=(xi+yj+zk)\cdot k=z\,$ and $\mathrm{curl}F\cdot n=-(i+j+k)\cdot(xi+yj+zk)=-(x+y+z)\,$

Therefore, $\iint_S(\mathrm{curl}F)\cdot n dS=\iint_R(\mathrm{curl}F)\cdot n\frac{dx dy}{n\cdot k}=\iint_R(-1)(x+y-z)\frac{dx dy}{z}=-\iint_R \frac{x+y+\sqrt{(1-x^2-y^2)}}{\sqrt{(1-x^2-y^2)}}dx dy\,$ --(3)

We now transform the double integral(3) into polar coordinates by writing $x=r\cos\theta,y=r\sin\theta,x^2+y^2=r^2,dx dy=r d\theta dr\,$ ,Again for the circular region,r varies from 0 to 1 and theta varies from 0 to $2\pi\,$ .Then (3) may be rewritten as

= $-\int_{\theta=0}^{2\pi}\int_{r=0}^{1} \frac{r\cos\theta+r\sin\theta+\sqrt{1-r^2}}{\sqrt{1-r^2}} r d\theta dr\,$

= $-\int_{\theta=0}^{2\pi}\int_{r=0}^{1}\frac{r^2}{\sqrt{1-r^2}}(\sin\theta+\cos\theta)d\theta dr=-\int_{0}^{1}\frac{r^2}{\sqrt{1-r^2}}(-\cos\theta+\sin\theta)_{0}^{2\pi}dr-\int_{r=0}^{1}r[\theta]_{0}^{2\pi}dr\,$

= $-\int_{0}^{1}\frac{r^2}{\sqrt{1-r^2}}(0)dr-\int_{0}^{1}r(2\pi)dr=0-2\pi[\frac{r^2}{2}]_{0}^{1}=-2\pi[\frac{1}{2}-0]=-\pi\,$

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