# Volume using triple integral with spherical and cylindrical coordinates [closed]

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).

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