Question: Use spherical coordinates to set up triple integrals for finding the volume of the region inside the sphere $x^2+y^2+z^2=4$ but outside the cylinder $x^2+y^2=1$
So question is pretty similar to my previous with a slight variation of the region. This is my answer:
$$\int _0^{2\pi }\int _{\frac{5\pi }{6}}^{\frac{\pi }{6}}\int _1^{\cos \phi \ }\rho \ ^2\sin \phi \ d\rho \, d\phi \ d\theta \ \ $$
Can anyone please please correct me? If my answer is correct then I'll remove this post, otherwise if there is any mistake, I'd be more than happy if someone can correct me!
Thanks in advance
Best Answer
Its almost correct! The only problem is the bounds for $\rho$. Since you want to describe $\rho$ in terms of the angle $\phi$ you can proceed as follows:
Fix two angles $\theta\in [0,2\pi),\ $$\phi\in [\pi/6,5\pi/6]$. Then for this two fixed angles take the unique point $(\rho,\theta,\phi)\equiv (x,y,z)$ which lies in the cylinder $x^2+y^2=1$. You want to calculate the radius $\rho$ of this point. Using the relations of $(x,y,z)$ in terms of $(\rho,\theta,\phi)$ and the equation $x^2+y^2=1$ we get \begin{align} (\rho \cos\theta \sin\phi)^2&+(\rho \sin\theta \sin \phi)^2=1\\ &\implies \rho^2\sin^2\phi=1\\ &\implies \rho=\frac{1}{|sin\phi|} \end{align} and since $\phi\in [\pi/6,5\pi/6]$ we eventually end up with $\rho=1/\sin\phi=\csc \phi$. Now , from that point and since the cylinder is inside the sphere of radius $2$ you move till you hit the sphere (for fixed $\theta,\phi$). This will happen when $\rho$ becomes $2$. So, the bounds for $\rho$ must be $\csc \phi\leq \rho\leq 2$.
Hence, the volume is $$\int_{0}^{2\pi}\int_{\pi/6}^{5\pi/6}\int_{\csc\phi}^{2}\rho^2\sin\phi\, d\rho d\phi d\theta$$