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I have the following standard problem I am not certain about how to solve. There are other problems in the book like this, I need some help with heuristics.

Let $S=(x,y): x^2+y^2=9, 0\le z\le5$, $F(x,y,z)=(2x,2y,2z)$. Determine the flux F through S.

Can someone please help me with this?

If I parametrize $x=9cos\alpha, y=9sin\alpha, z=z$, is $dS=9dzd\alpha$? Also, how do I find the unit vector? Is it $n=\frac{xi+yj}{9}$? These confuse me

Thanks in advance!

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1 Answer

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For $3$-dimensional problems, getting the vector areal element is easy because we have the cross product available to us. First, parameterize the surface in terms of two variables. You have chosen $\vec r=\langle3\cos\theta,3\sin\theta,z\rangle$ along the surface. I have fixed your value of $r$ because the equation is $r^2=9$, not $r=9$. Now we find the differential of the of the position vector: $$d\vec r=\langle-3\sin\theta,3\cos\theta,0\rangle\,d\theta+\langle0,0,1\rangle\,dz$$ These two differential vectors point long the surface, the first one the magnitude and direction of the vector change in position along the surface when $\vec r$ goes from $\vec r(\theta,z)$ to $\vec r(\theta+d\theta,z)$ and the second carries the magnitude and direction of the vector change in position vector when we go fraom $\vec r(\theta,z)$ to $\vec r(\theta,z+dz)$. thus the cross product gives the a vector normal to the surface (because both vectors are parallel to the surface) and of area equal to the corresponding parallelogram on an imaginary grid drawn in curves of constant $\theta$ and $z$ along the surface. Thus $$\begin{align}d^2\vec A&=\pm\langle-3\sin\theta,3\cos\theta,0\rangle\,d\theta\times\langle0,0,1\rangle\,dz\\ &=\pm\langle3\cos\theta,3\sin\theta,0\rangle\,d\theta\,dz\\ &=\langle3\cos\theta,3\sin\theta,0\rangle\,d\theta\,dz\end{align}$$ Because we want the areal element to point out of the cylinder. This is so dead simple in $3$-d that I don't know why they would ever teach anything else, but they do :(

Now we have to parameterize the vector field in terms of the same variables $$\vec F=\langle2x,2y,2z\rangle=\langle6\cos\theta,6\sin\theta,2z\rangle$$ So the flux out of the curved surface is $$\begin{align}\Phi&=\int\vec F\cdot d^2\vec A=\int_0^{2\pi}\int_0^5\langle6\cos\theta,6\sin\theta,2z\rangle\cdot\langle3\cos\theta,3\sin\theta,0\rangle\,dz\,d\theta\\ &=\int_0^{2\pi}\int_0^518\,dz\,d\theta=(18)(2\pi)(5)=180\pi\end{align}$$

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