Volume between a sphere and a plane in Cartesian system 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 $dxdydz$

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?
