# Finding the volume inside a sphere above a cone

I'm having some problems finding the volume inside a sphere of radius 2 and above the cone $\sqrt3z = \sqrt{x^2 + y^2}$.

I tried integrating using spherical coordinates and got $\int_0^{2\pi}\int_0^{\pi/2}\int_1^2 \rho^2 sin\phi$ $d\rho$ $d\phi$ $d\theta$, which gave me $14/3\pi$, but the correct answer is $8/3 \pi$, anyone knows what might have gone wrong or what is the correct way to solve this problem?

Much appreciated!

It looks like you have an incorrect bound on the angle $\phi$. The cone in question is the upper half of $3z^2 = r^2$, so at height $z = 1$, for example, the radius of the cone is $3$. Drawing an appropriate right triangle will allow you to find the angle between your cone's side and the $z$-axis. This will be the upper bound on $\phi$ in your integral.

Happy calculating!

• The radius is $\sqrt3$ you mean? Also, are the limits for $\rho$ wrong? I have a feeling that it should be some expression in terms of $\phi$ and $\theta$ instead of just constants, but I'm not sure how to get those expressions? – inggumnator Jan 14 '15 at 4:01
• Ooh, yes. The lower bound on $\rho$ should be zero since you want the portion inside a sphere. (Using bounds 1 and 2 give you the volume of the cone within a spherical 'shell'.) In this case you have radial symmetry so your bounds won't actually depend on $\theta$ or $\phi$ (you travel from $\rho = 0$ out to $\rho = 2$ when integrating along each choice of $\theta$ and $\phi$). – Titus Jan 14 '15 at 4:05

$$V= \int_{0}^{2\pi} \int_{0}^{\sqrt{3}} (Z_{top} - Z_{bottom})rdrd\theta$$
$$V= \int_{0}^{2\pi} \int_{0}^{\sqrt{3}} \left(\sqrt{4-r^2} - \frac{r}{\sqrt{3}}\right)rdrd\theta$$
For the first integral make a substitution $(4-r^2) = u$
And if you evaluate, you get $V = \frac{8\pi}{3}$