Volume of sum of a cube and a sphere

Question

Let $$C=\{(x, y, z)\in \mathbb{R}^3:|x|\leq1, |y|\leq 1, |z|\leq 1\}$$ be a cube and $$S=\{(x, y, z)\in \mathbb{R}^3:x^2+y^2+z^2\leq 1\}$$ be a sphere. Let us define $$C+S=\{c+s:c\in C, s\in S\}.$$Find the volume of $C+S$.

The main requirement is to determine the limits of $x, y$ and $z$ for the set $C+S$. After evaluating them, I get $$-2\leq x\leq 2$$ $$-1-\sqrt{1-x^2}\leq y\leq 1+\sqrt{1-x^2}$$ $$-1-\sqrt{1-x^2-y^2}\leq z\leq 1+\sqrt{1-x^2-y^2}.$$ This means the volume of $C+S$ is $$\int_{-2}^2\int_{-1-\sqrt{1-x^2}}^{1+\sqrt{1-x^2}}\int_{-1-\sqrt{1-x^2-y^2}}^{1+\sqrt{1-x^2-y^2}}dz\, dy\, dx.$$ Is this the correct approach? Please help. Also, if this is the correct approach, then how to evaluate the integral $$ 2\int_{-2}^2\int_{-1-\sqrt{1-x^2}}^{1+\sqrt{1-x^2}}\sqrt{1-x^2-y^2} dy\, dx.$$

Answer 1

Intuitively $C+S$ consists of

• Eight eighth-spheres of radius $1$ (purple), with total volume $\frac43\pi$
• $12$ quarter-cylinders of height $2$ and radius $1$ (blue), with total volume $3\cdot2\pi=6\pi$
• Six $2×2×1$ cuboids (white) with total volume $24$
• a central cube (pink) of side $2$ which is $C$ and has volume $8$

Hence the total volume is $32+\frac{22}3\pi$.