Volume between a sphere and a cone

Question

Find the volume between $r=R\,$ (sphere) and $\theta=\alpha\,$ (cone) , for $\theta$ and $r $ constants such that - $0 < \theta < \dfrac{\pi}{2}$

I am sorry for the question being basic, but i couldn't find similar questions on Math Exchange.

I can't figure what my integration limits suppose to be in order to solve this in a spherical coordinate system -

$\phi$ is obviously from $0$ to $2\pi$, $\theta$ from $0$ to $\alpha$ but how can i bound $r$ ?

Thank you !

Answer 1

$r$ should vary between $0$ and $R$.

The volume you're looking for is

$$V= \int_0^R \int_0^\alpha \int_0^{2 \pi} r^2 \sin \theta d \phi \ d\theta \ dr= \frac{2}{3} \pi R^3 (1- \cos \alpha)$$

And by the way for $\alpha = \frac{\pi}{2}$ you can verify that you get the volume of half a sphere.

Answer 2

The solution can be expressed in spherical coordinates as

$$ \begin{align} V &=\iiint~dV\\ &=\int_0^{\alpha}\sin\phi~d\phi~\int_0^{2\pi}d\theta~\int_0^Rr^2dr\\ &=(1-\cos\alpha)\cdot2\pi\cdot \frac{R^3}{3}\\ &=\frac{2}{3}\pi R^3(1-\cos\alpha) \end{align} $$