Can You Solve the Divergence Theorem for a Cylindrical Vector Field?

Baumer8993
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Homework Statement


Verify the divergence theorem by computing both integrals for the vector field
F = <x^3, y^3, z^2> over a cylindrical region define by x^2+y^2 ≤ 9.


Homework Equations


Divergence Theorem, and Flux Integrals.


The Attempt at a Solution


I did the divergence theorem, and got 279 pi for my answer. I did the integral in cylindrical, and regular with the same answer so that I know is correct. I know I need to break this up into three different surfaces. I can do the top, and the bottom easily since N = <0, 0, 1> for the top, and N = < 0, 0, -1> for the bottom, then I just plug in the z value for it.

The side of the cylinder is giving me trouble. I have looked up ways to do it online, but they make no sense, or are disorganized. Where do I start, and then where do I go from there?
 
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You haven't completely stated the problem. You say it is over the cylidrical region x^2+ y^2\le 9 but don't say what the ends are. What values of z?

You want to integrate over the surface of the cyinder. Any point on the circle x^2+ y^2= 9 can be written (3cos(\theta), 3sin(\theta))[/tiex] with \theta going from 0 to 2\pi. They cylinder has axis along the z-axis so we can take any point on the cylinder as given by \vec{r}(\theta, z)= 3cos(\theta)\vec{i}+ 3sin(\theta)\vec{j}+ z\vec{k}.<br /> <br /> The \vec{r}_\theta= -3sin(\theta)\vec{i}+ 3cos(\theta)\vec{j}and \vec{r}_z= \vec{k} and the cross product of those vectors gives the vector differential of surface area, (3cos(\theta)\vec{i}+ 3 sin(\theta)\vec{j})d\theta dz.&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; The two ends will have vector differential \vec{k} drd\theta and -\vec{k}drd\theta with r going from 0 to 3 and \theta going from 0 to 2\pi
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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