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.
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1$\begingroup$ What's the center of the sphere? $\endgroup$– Mhenni BenghorbalNov 20, 2013 at 7:39
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$\begingroup$ Related problems. $\endgroup$– Mhenni BenghorbalNov 20, 2013 at 8:07
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1 Answer
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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?
answered Nov 20, 2013 at 7:51
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2$\begingroup$ @Artemisia: Just noting: "don't forget considering the proper Jacobian in the triple integrals. $\endgroup$– MikasaNov 20, 2013 at 8:07
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$\begingroup$ 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. $\endgroup$ Nov 20, 2013 at 11:34
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$\begingroup$ 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