What is the solution to this Divergence Theorem homework problem?

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SUMMARY

The solution to the Divergence Theorem homework problem involves evaluating the integral of the vector field F = 5y + 4z + 7x over a specified region in the first octant. The correct setup for the triple integral is int(theta from 0 to pi/2) int(r from 0 to sqrt(3)) int(z from 1 to 4 - r^2) (5r^2sin(theta) + 4rz + 7r^2cos(theta)) dz dr dtheta. The initial error in the setup was using an incorrect upper limit for r, which should be sqrt(3) instead of 2. The final evaluation of the integral yields the correct result.

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Homework Statement


evaluate https://instruct.math.lsa.umich.edu/webwork2_files/tmp/equations/71/7816ab9562fbe29a133b96799ed5521.png if https://instruct.math.lsa.umich.edu/webwork2_files/tmp/equations/65/11ed69ea372626e9c4cee674c8dc6f1.png and S is the surface of the region in the first octant bounded by x = 0, y = 0, below by z = 1, and above by https://instruct.math.lsa.umich.edu/webwork2_files/tmp/equations/70/dd75ed46cf7f510c406a2b2e8cd0cd1.png


Homework Equations





The Attempt at a Solution


I used the divergence of F=5y+4z+7x.
My integral was
int(theta from 0 to pi/2)int(r from 0 to 2)int(z from 1 to 4-r^2) (5r^2sin(theta)+4rz+7r^2cos(theta)) dzdrdtheta.
I get 26pi/3+96/5, but that's not the right answer. Is my setup wrong or am I evaluating it wrong?
 
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You set up is correct. I haven't checked the evaluation of the integral.
 
Thanks for the help, but I figured it out and my setup wasn't correct. At z=1 which is the base of the region, r goes to sqrt(3) not 2. So the r integral goes from 0 to sqrt(3).
 

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