Calculating volume of a set $K=\{(x,y,z)^T \in \mathbb{R}^3:x^2+y^2+z^2 \le 4, x^2+y^2 \le 2x\}$

Question

I want to to calculcate the volume, that is $\lambda_3$, of $K=\{(x,y,z)^T \in \mathbb{R}^3:x^2+y^2+z^2 \le 4, x^2+y^2 \le 2x\}$.

I guess I need to evaluate a triple integral for this but I don't know how to this task.

Answer 1

In this type of problems, try to make a sketch of the volume. The first inequality is the interior of the sphere with radius $2$, centered at the origin.

If you move $2x$ to the other side in the second inequality, then complete the square, you get that this is the interior of the cylinder of radius $1$, with axis parallel to $z$, and centered on $x=1$.

So one way to set up the calculation is to look at the $z=0$ plane. The integration range in this plane is the circle with center at $x=1$, $y=0$ and radius $1$. So the limits for $x$ are $0$ to $2$, and the limits for $y$ are from $-\sqrt{2x-x^2}$ to $+\sqrt{2x-x^2}$. The limits for $z$ you get from the first inequality, $z$ varies between $\pm\sqrt{4-y^2-x^2}$. Therefore $$V=\int_0^2dx\int_{-\sqrt{2x-x^2}}^{\sqrt{2x-x^2}}dy\int_{-\sqrt{4-y^2-x^2}}^\sqrt{4-y^2-x^2}dz$$