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Use the divergence theorem to calculate $\int \int F\cdot dS$ where $F=<x^3,y^3,4z^3>$ and $S$ is the sphere $x^2+y^2+z^2=25$ oriented by the outward normal.

I have found that $div(F)=<3x^2,3y^2,12z^2>$ and set up the an integral to solve using the formula $\int \int F\cdot dS=\int \int \int div(F) dv$.

As such, I converted to spherical coordinates and simplified to the expression(including the $\rho^2sin(\phi)$ factor):

$\int \int \int3\rho^4sin^3(\phi)+12\rho^4sin(\phi)cos^2(\phi)d\rho d\phi d\theta$

with the bounds$\quad0\le \rho \le 5,\quad0 \le\phi \le \pi, \quad 0\le\theta\le 2\pi$

My result came out as $47123.9$ which is apparently not correct. Anyone want to help me figure out what I'm doing wrong? Thanks!

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  • $\begingroup$ The correct answer is indeed $15000 \pi$. $\endgroup$ – Maxim Oct 26 '18 at 14:54
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In spherical polar coordinates $3x^2 + 3y^2 + 12z^2 = 3(x^2 + y^2 + z^2) + 9 z^2$ and $x^2 + y^2 + z^2 = r^2$. This means you have to evaluate the integral $\int \int \int (3 r^2 + 9z^2) r^2dr \sin \theta d\theta d\phi$. After using $z = r \cos \theta$ you eventually need to determine $\int \int \int (3 r^2 + 9r^2 \cos^2 \theta) r^2dr \sin \theta d\theta d\phi$.

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  • $\begingroup$ Using this technique, I an still getting the same result, which was incorrect before. $\endgroup$ – Mike9060 Apr 21 '16 at 21:04

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