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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

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  • $\begingroup$ You probably didn't have the correct bounds for the just the first octant. $\endgroup$
    – Ninad Munshi
    Feb 14, 2021 at 6:36
  • $\begingroup$ 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. $\endgroup$
    – Ninad Munshi
    Feb 14, 2021 at 6:39

1 Answer 1

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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 $\displaystyle \frac{1}{256}$.

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  • $\begingroup$ Thank you @Math Lover $\endgroup$
    – simon
    Feb 14, 2021 at 7:26
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    $\begingroup$ 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. $\endgroup$
    – Math Lover
    Feb 14, 2021 at 7:33
  • $\begingroup$ 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? $\endgroup$
    – simon
    Feb 14, 2021 at 7:51
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    $\begingroup$ 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 $\tan^2 \phi = \frac{1}{3}$. $\endgroup$
    – Math Lover
    Feb 14, 2021 at 7:58
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    $\begingroup$ 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. $\endgroup$
    – Math Lover
    Feb 14, 2021 at 8:08

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