Find the volume of the region inside both the sphere $x^2+y^2+z^2=4$ and the cylinder $x^2+y^2=1$

Question

Find the volume of the region inside both the sphere $x^2+y^2+z^2=4$ and the cylinder $x^2+y^2=1$
$$2\int_0^{2\pi}\int_0^{\frac{\pi} 6}\int_0^{2}1.r^2\sin \varphi dr d\varphi d\theta + \int_0^{2\pi}\int_{\frac{\pi} 6}^{\frac{5\pi} 6}\int_0^{\frac 1 {\sin \varphi}}1.r^2\sin \varphi dr d\varphi d\theta $$

This question is in the lecture notes but I didn't understand how the answer is obtained. Could someone help?

Answer 1

Note that

• the intersection of the cilinder and the sphere is for $z=\pm \sqrt 3$ that is $\varphi=\frac{\pi}6$ and $\varphi=\frac{5\pi}6$

• the first integral is twice the "ice cream" upper part of the sphere (in spherical coordinates)

• the second integral is the residual part inside the cylinder (in spherical coordinates)

• in the second integral $r(\varphi)=\frac1{\sin \varphi}$

Answer 2

You obtain a spherical ring by drilling out a hole of the sphere with the cylinders radius. It is well known that its volume solely depends on the height of the ring, namely $V=\frac{\pi}{6}h^3$, see Remaining volume of a sphere.

Now subtract that volume from the volume of the sphere. In our case $h=\sqrt3$, hence the volume we’re looking for is $$\frac43\pi\cdot2^3-\frac\pi6\sqrt{3}^3.$$