Volume between a sphere and a plane in Cartesian system

Question

How can I calculate the volume between the sphere $r=R$ and the plane $z=R/2$ (above the plane and below the sphere) in Cartesian system?

edit - using triple integral $dxdydz$

Answer 1

Using a triple integral is not a good idea..... but can be a good exercise!

So, as a first step note that the projection on the plane $x-y$ of the itersection of the sphere of radius $R$, centered at the origin, with the plane of equation $z=R/2$ is a circle of equation $x^2+y^2=\frac{3}{4}R^2$.

This means that the limits of integration for $y$ are: $$ -\sqrt{\frac{3}{4}R^2-x^2}<y<\sqrt{\frac{3}{4}R^2-x^2} $$ and $y$ is a real number if $x$ has limits: $$ -\frac{\sqrt{3}}{2}R<x<\frac{\sqrt{3}}{2}R $$

The equation of the sphere in cartesian coordinates is $x^2+y^2+z^2=R^2$, so we have the limits of integration for $z$:

$$ \frac{R}{2}<z<\sqrt{R^2-x^2-y^2} $$

Putting all toghether we have that the volume is: $$ \int_{-\frac{\sqrt{3}}{2}R}^{\frac{\sqrt{3}}{2}R} \int_{-\sqrt{\frac{3}{4}R^2-x^2}}^{\sqrt{\frac{3}{4}R^2-x^2}} \int _{\frac{R}{2}}^{\sqrt{R^2-x^2-y^2}} dz dy dx $$

This integral is difficult and requires a trigonometric substitution. A simpler solution is to use cylindrical coordinates. Can you find what the integral becomes in this case?