The following function is given: $$\iiint_{x^2+y^2+z^2\leq z} \sqrt{x^2+y^2+z^2}dx\,dy\,dz$$ And I have to calculate this integral using spherical coordinates. The substitutions are standard, I think, but I am having a problem with the limits. $$0\leq\phi\leq\pi$$$$0\leq\theta\leq2\pi$$ are the limits for the angles. I am not able to determine the limits for $\rho$ defined as $$\rho=\sqrt{x^2+y^2+z^2}$$ I tried it with $\rho\cos(\phi)$ as the upper limit but it didn't work.


1 Answer 1


You're given the region $x^2+y^2+z^2 \leq z$, which by definition is equivalent to $\rho^2 \leq \rho\sin\phi$. You can assume $\rho \geq 0$, so this simplifies to $0 \leq \rho \leq \sin\phi$.

Is that enough for you to figure out the problem?

  • 2
    $\begingroup$ @Artemisia: Just noting: "don't forget considering the proper Jacobian in the triple integrals. $\endgroup$
    – Mikasa
    Nov 20, 2013 at 8:07
  • $\begingroup$ Are you sure that it is $\sin\phi$? I am not getting the answer using that. And yes I did include the Jacobian. I am getting $\pi/5$ but the system says that's incorrect. $\endgroup$
    – Artemisia
    Nov 20, 2013 at 11:34
  • $\begingroup$ Well my logic appears to be sound as far as I can tell. When I work it out, I don't get $\pi/5$, so you must have an arithmetic error. I don't want to straight out give you the answer because it sounds like an assignment, but to help you - there is indeed a 5 in the denominator of the answer. Are you sure you're using the Jacobian properly, $\rho^2\sin\phi$? $\endgroup$ Nov 20, 2013 at 19:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .