# compute $\iiint_Kxyz\ dxdydz$

The question is: $$\iiint_Kxyz\ dxdydz\quad k:=\{(x,y,z):x^2+y^2+z^2\leq1, \ \ x^2+y^2\leq z^2\leq 3(x^2+y^2), \ x,y,z\geq 0\}$$

Here how i have tried to solve this: $$\iint_{x^2+y^2\leq1}\int_{\sqrt{x^2+y^2}}^{\sqrt{3(x^2+y^2)}}xyz \ dz\ dxdy=\frac{1}{2}\iint xy\left(3(x^2+y^2)-(x^2+y^2)\right)=...=0$$ But the answer that i got is zero which is obviously wrong what is wrong with my solution? any suggestion would be great, Thanks

• You probably didn't have the correct bounds for the just the first octant. Feb 14, 2021 at 6:36
• More importantly, the way you set up your bounds was incorrect. If you were going to do $dz$ first, this would need to be two separate integrals. Feb 14, 2021 at 6:39

Basically we are trying to find the volume of the sphere $$x^2+y^2+z^2 \leq 1$$ between two cones $$z^2 = x^2 + y^2$$ and $$z^2 = 3(x^2+y^2)$$.

So the bounds are much easier to set up in spherical coordinates.

$$x = \rho \cos \theta \sin \phi, y = \rho \sin \theta \sin \phi, z = \rho \cos \phi$$

$$0 \leq \rho \leq 1, 0 \leq \theta \leq \frac{\pi}{2}$$ are obvious as we are in first octant and radius of the sphere is $$1$$.

Now to find the bounds of $$\phi$$, we observe that

$$x^2 + y^2 \leq z^2 \leq 3(x^2+y^2)$$

plugging in $$x,y,z$$, $$\frac{1}{\sqrt3} \leq \tan \phi \leq 1 \implies \frac{\pi}{6} \leq \phi \leq \frac{\pi}{4}$$

$$xyz = \rho^3 \cos\theta \sin \theta \sin^2\phi \cos\phi$$

So the integral becomes,

$$\displaystyle \int_{0}^{\pi/2} \int_{\pi/6}^{\pi/4} \int_0^1 \rho^5 \cos\theta \sin \theta \sin^3\phi \cos\phi \ d\rho \ d\phi \ d\theta$$

I did not do the integral by hand but WolframAlpha shows the result as $$\displaystyle \frac{1}{256}$$.

• Thank you @Math Lover Feb 14, 2021 at 7:26
• You are welcome. I would suggest you to visualize this in Geogebra 3D. Are you comfortable with spherical coordinates? Otherwise it can be set up in cylindrical / cartesian too. Feb 14, 2021 at 7:33
• Yes thank you again, just one little thing, in order to find $\phi$ as you know we need to divide by r and $\cos \phi$ i wonder they do not become zero? Feb 14, 2021 at 7:51
• Let's take one of them - $3(x^2+y^2) = z^2 \implies 3 \rho^2 \sin^2 \phi = \rho^2 \cos^2 \phi$. That gives you $\tan^2 \phi = \frac{1}{3}$. Feb 14, 2021 at 7:58
• Yes it is because this is to just find intersection of cones with the sphere for our integral bounds and in any case, $\rho$ is not zero at the intersection, we know that. Feb 14, 2021 at 8:08