A solid sphere $x^2+y^2+z^2 \le 1$ is cut into two parts by a plane $z=\frac{1}{2}$.

Question

A solid sphere $x^2+y^2+z^2 \le 1$ is cut into two parts by a plane $z=\frac{1}{2}$. Find the volume of the smaller part. I have no idea how to approach this. It will be helpful if you give me the general way to approach these problems.

Answer 1

HINT

By cylindrical coordinates we have

$$V=\int_0^{2\pi}d\theta\int_{\frac12}^1 dz\int_0^{\sqrt{1-z^2}}r\,dr=\pi\int_{\frac12}^1 (1-z^2)dz$$

As an alternative we could also use spherical coordinates but I think that cylindrical coordinates are more effective to use here.

If you are not forced to use triple integral, as noticed by B. Goddard, you may refer to the formula for Volumes of Solids of Revolution that is

$$\int_a^b \pi (f(z))^2 dz$$

obtaing of course the same result.

Answer 2

This is a Calc I problem. Rotate the region under the curve $y=\sqrt{1-x^2}$ from $x=1/2$ to $1$ about the $x$-axis.

$$\int_{1/2}^1 \pi \left(\sqrt{1-x^2}\right)^2 \; dx.$$

Answer 3

Volume of the smaller part (a spherical cup) $$V=\frac{\pi h^2}{3}(3r-h)$$ Note that here $r=1$ and $h=r-(1/2)=1/2$