I had to solve this triple integral and I tried to solve by cylindrical and spherical coordinates but couldn't get anywhere. I was hoping someone could help me in this problem.

Solve $\int \int \int _{E} \frac{1}{x^2+y^2+z^2 }dV, E=\{(x,y,z) \epsilon \mathbb{R}^3 : z\geq \sqrt{\frac{x^2+y^2}{3}}, 2z\leq x^2+y^2+z^2 \leq 9\} $


You can do this in both cylindrical and spherical coordinates but it is more straightforward in spherical coordinates.

Based on inequalities, the region $E$ we need to integrate over is the volume of cone $z \geq \sqrt{\frac{x^2+y^2}{3}} \ $ inside the sphere $x^2 + y^2 + z^2 \leq 9 \ $ but outside the other sphere $x^2+y^2+z^2 \geq 2z$.

In spherical coordinates, $x = \rho \cos \theta \sin \phi, y = \rho \sin \theta \sin \phi, z = \rho \cos \phi, x^2 + y^2 + z^2 = \rho^2$

So the maximum polar angle $\phi$ from the equation of the cone,

$\rho \cos \phi = \frac{\rho \sin \phi}{\sqrt3} \implies \phi = \frac{\pi}{3}$

The equation of the second sphere can be re-written as

$\rho^2 = 2 \rho \cos \phi \implies \rho = 2 \cos \phi, 0 \leq \phi \leq \frac{\pi}{2}$

So the integral is $\displaystyle \int_0^{2\pi} \int_0^{\pi/3} \int_{2\cos \phi}^3 \sin \phi \ d\rho \ d\phi \ d\theta$

(Please note that the integrand is $\frac{1}{\rho^2}$ which cancels out with $\rho^2$ in the Jacobian).

  • $\begingroup$ Can you confirm that the result is zero? $\endgroup$ – Raffaele Jan 16 at 21:43
  • $\begingroup$ No @Raffaele, I had not checked the value earlier. I just did. It is $\frac{3 \pi}{2}$. $\endgroup$ – Math Lover Jan 16 at 22:21
  • $\begingroup$ Thank you again! $\endgroup$ – Vinicius Pansini Jan 16 at 23:11

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