Finding the volume between a cone and a sphere

Question

I have to find the volume between the sphere $x^2+y^2+z^2=1$ and below the cone $z=\sqrt{x^2+y^2}$ using Spherical Coordinates.

Here is what I have so far:

Transforming the cone part gives:

$\begin{align*} \sqrt{x^2+y^2} &= \sqrt{r^2\cos{\theta}^2\sin{\phi}^2+r^2\sin{\theta}^2\sin{\phi}^2}\\ &=\sqrt{r^2\sin{\phi}^2(\cos{\theta^2}+\sin{\theta}^2)}\\ &=r\sin{\phi} \end{align*}$

I know the shape of intersection is a circle with $r = \frac{1}{4}$.

I am pretty sure that as far as the limits of integration will go, $0\leq\theta\leq 2\pi$, but I have no idea where to go from here.

How can I set up the integral?

Answer 1

HINT:

This is not a proper answer using spherical coordinates so giving as hint for answer, hope details are easy to fill in. Area of spherical cap is area of equatorial circle multiplied by cap height.

$$ \frac{Area_{spherical cap}}{ 4 \pi R^2} \cdot (\frac43 \pi R^3) $$

$$ = 2\pi R^2 (1- \frac{1}{\sqrt2})\cdot \frac{R}{3} $$

Maximum radius of cone is $ \frac{R}{\sqrt2}.$

Answer 2

The most advisable is to use the Change of Variables Theorem, in that sense, we have the integral of the region is:

$$\int_{0}^{\frac{1}{\sqrt{2}}}\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{4}}r^{2}\sin \left(\phi \right) d\phi d\theta dr=\frac{1}{6}\left(\sqrt{2}-1\right)\pi$$ The calculations of the integral is left as an exercise.