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.
Best Answer
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, $$0\leq\rho\leq\frac h{\cos\phi}.$$