We were given this exercise in class to take home but I am a bit confused with it. If anyone could help I would appreciate it.

Let $C$ be a conical solid bounded above by $z=h$ and below by the cone $z=k\cdot\sqrt{x^2+y^2}$, $k>0$. We are supposed to find the triple integral for the volume of the cone in spherical coordinates.

$$V = \int\int\int \rho^2\sin(\theta)\,d\rho\,d\phi\,d\theta $$

I am a bit confused at what the boundaries will be. I know $ 0\leq\theta\leq2\pi$ however I am confused at what the boundaries for $\phi$ and $\rho$ will be. If someone could shed some light on this I would really appreciate it.


There are a few things we need to keep in mind. First of all, $$x=\rho\sin\phi\cos\theta\\y=\rho\sin\phi\sin\theta\\z=\rho\cos\phi.$$ Then $$\sqrt{x^2+y^2}=\rho\sin\phi$$ Hence, the surface $$z=k\sqrt{x^2+y^2}$$ can be rewritten spherically as $$\rho\cos\phi=k\rho\sin\phi,$$ or $$\phi=\tan^{-1}(1/k)$$ (since we're not below the $xy$-plane). The positive $z$-axis corresponds to $\phi=0,$ so we have $$0\le\phi\le\tan^{-1}(1/k).$$

Note that the surface $z=k\sqrt{x^2+y^2}$ ended up being independent of $\rho$. Hence, the only boundary on $\rho$ aside from the origin ($\rho=0$) is the surface $z=h$, which we can rewrite spherically as $$\rho=\frac h{\cos\phi}.$$ Thus, $$0\leq\rho\leq\frac h{\cos\phi}.$$

  • $\begingroup$ Ok I understand all of this... Now then is the integral I have written below in the answer section the correct integral then? I am just a bit confused because normally when I am doing multiple integration like this my boundaries are not so complicated. I guess our prof gave this to us as a thinking question. $\endgroup$ – user68203 Mar 27 '13 at 5:43
  • $\begingroup$ You're very close. First of all, you lost your $\sin\theta$ term. Also, note that when you integrate with respect to $\rho$, you'll get another terms dependent on $\phi$, so you can't pass let the integral with respect to $\rho$ "pass through" the integral with respect to $\phi$. It should be $$\int_0^{2\pi}\sin\theta\,d\theta\int_0^{\tan^{-1}k}\cos\phi\int_0^{h/\sin\phi}\rho^2\,d\rho\,d\phi.$$ This is an excellent question to get you used to spherical coordinates, and seeing how they work. $\endgroup$ – Cameron Buie Mar 27 '13 at 14:56
  • $\begingroup$ Now I just got confused. Earlier I noticed you used spherical coordinates $$x=\rho cos\phi cos\theta$$ $$y=\rho cos\phi sin\theta$$ $$z=\rho sin\phi$$ Normally I use $$x=\rho sin\phi cos\theta$$ $$y=\rho sin\phi sin\theta$$ $$ z=\rho cos\phi$$ I know the latter corresponds to $\rho^2 sin\phi d\rho d\phi d\theta$ from my text book, however your coordinates I thought when I checked them using change of variables corresponded to $\rho^2 cos\phi d\rho d\phi d\theta$ . Can you confirm this? Then Now I am just confused at where your extra $sin\theta$ or $cos\phi$ comes from... $\endgroup$ – user68203 Mar 27 '13 at 15:20
  • $\begingroup$ Ah! You're correct. I swapped $\sin\phi$ and $\cos\phi$ in my conversions by mistake, and it should be a $\sin\phi$ instead of $\sin\theta$ (which you put in your question). My answer has been corrected, and the integral should then be $$\int_0^{2\pi}\int_0^{\tan^{-1}(1/k)}\int_0^{h/\cos\phi}\rho^2\sin\phi\,d\rho\,d\phi\,d\theta=\int_0^{2\pi}\,d\theta\int_0^{\tan^{-1}(1/k)}\sin\phi\int_0^{h/cos\phi}‌​\rho^2\,d\rho\,d\phi.$$ $\endgroup$ – Cameron Buie Mar 27 '13 at 15:31
  • $\begingroup$ OK awesome. Thanks so much for your help! $\endgroup$ – user68203 Mar 27 '13 at 15:55

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