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Evaluate the surface integral $ \int_{S}\int \vec{F} \cdot \vec{n}\, dS,$ with the vector field $ \vec{F} = zx\vec{i} + xy\vec{j} + yz\vec{k} \ $. $S$ is the closed surface composed of a portion of the cylinder $ x^2 + y^2 = R^2 $ that lies in the first octant, and portions of the planes $ x=0, y =0, z = 0\,\,\text{and}\,\, z = H $. $\vec{n}$ is the outward unit normal vector.

Attempt: I said $S$ consisted of the five surfaces $ S_1, S_2, S_3, S_4 $ and $ S_5$ $S_1 $ being the portion of the cylinder, $S_2$ being where the plane $ z=0$ cuts the cylinder and similarly, $ S_3, S_4 ,S_5 $ being where the planes $ x = 0, y = 0 $ and $z = H $ cut the cylinder.

For $ S_2 $, the normal vector points in the -k direction. so the required integral over $S_2$ is: $$ \int_{0}^{R} \int_{0}^{\sqrt{R^2-x^2}} -yz\,dy\,dx $$ Am I correct? I think for the surface $ S_5$ the only thing that would change in the above would be that the unit normal vector points in the positive k direction?

I need some guidance on how to set up the integrals for the rest of the surfaces. I tried $ \int_{0}^{R} \int_{0}^{H} -xy\,dz\,dx $ for the y = 0 plane intersection with the cylinder, but I am not sure if this is correct.

Any advice on how to tackle the remaining surfaces would be very helpful. Many thanks.

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up vote 1 down vote accepted

On the $x=0$ face $S_3$ one has $n=(-1,0,0)$ and $F\cdot n=0$.

On the $y=0$ face $S_4$ one has $n=(0,-1,0)$ and $F\cdot n=0$.

On the $z=0$ face $S_2$ one has $n=(0,0,-1)$ and $F\cdot n=0$.

It follows that the integrals over these faces are zero.

On the $z=H$ face $S_5$ one has $n=(0,0,1)$ and $F\cdot n=Hy$. Using polar coordinates in the $(x,y)$-plane gives $y=r\sin\phi$, $\ {\rm d}\omega={\rm d}(x,y)=r{\rm d}(r,\phi)$ and therefore $$\int\nolimits_{S_5}F\cdot n\ {\rm d}\omega=\int_0^R\int_0^{\pi/2} H r^2 \sin\phi \ d\phi\ dr=\ldots\quad.$$

On the $x^2+y^2=R^2$ face $S_1$ one has $$x=R\cos\phi,\quad y=R\sin\phi,\quad z=z,\quad n=(\cos\phi,\sin\phi,0)$$ and therefore $$F\cdot n=Rz\cos^2\phi+R^2\cos\phi\sin^2\phi\ .$$ Furthermore ${\rm d}\omega=R\ {\rm d}(z,\phi)$. It follows that $$\int\nolimits_{S_1}F\cdot n\ {\rm d}\omega=R\ \int_0^H\int_0^{\pi/2}(Rz\cos^2\phi+R^2\cos\phi\sin^2\phi)\ d\phi\ dz=\ldots\quad.$$

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Use the Divergence Theorem.

$\text{div}(F) = x + y + z$

The volume integral is $$\int_{0}^{2\pi} \int_{0}^{R} \int_{0}^{H} (r\cos\theta + r\sin\theta + z)r dz dr d\theta$$

I leave the integration to you

In the end you should get $$\frac{1}{2}\pi HR^2$$

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Many thanks for your reply. We haven't done the divergence theorem yet and the above is a set question. Can you give me hints on how to do it using surface integral techniques? – CAF Nov 3 '12 at 18:47

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