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I have to solve this integral $\int_{x^2+y^2+z^2-2x\le0} \frac{x}{x^2+y^2+z^2}dxdydz$.
So, saw the denominator of the funcion I try substitution with polar ball coordinates, so the integral becomes:
$\int_0^{2\pi}\int_0^\pi\int_0^1\frac{\rho\sin\psi cos\theta}{\rho^2}\rho^2sin\psi d\theta d\psi d\rho$
but the result of this integral is 0, when i know that the result of the first integral is $\pi$. Can you tell me where I'm wrong.

1 Answers 1

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$x^2+y^2+z^2-2x\le0$ so $0\le \rho \le 2cos(\theta)sin(\phi)$

When $z=0$

Note that $-\frac {\pi}{2} \le \theta \le \frac{\pi}{2}$

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    So trying $\int_0^{2\pi}\int_0^{\pi}\int_0^{2cos\theta\sin\psi} \rho(sin\psi)^2cos\theta d\theta d\psi d\rho$ but still not work. – 2017-01-10
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    Its because $\frac{-\pi}{2} \le \theta \le \frac{\pi}{2}$ – 2017-01-10
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    ok. I tryed it, and the result is right. But I still not understanding something. Can you explain me, why $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$ ? – 2017-01-10
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    Its because the sphere , when $z=0 $, is only in the first and fourth quandrant would be $0 \le \theta \le 2\pi $ if the sphere was in all quadrants. You can imagine the +x-axis as $0$, +y-axis as $\frac{\pi}{2}$, -x-axis as $\pi$ and the -y-axis as $\frac{3\pi}{2}$. Just use the trigonometric circle. – 2017-01-10