Compute $\iiint_G \frac{xyz}{(a^2 + x^2 + y^2 + z^2)^3}dxdydz$.

Question

How to compute this integral ? $$\iiint_G \frac{xyz}{(a^2 + x^2 + y^2 + z^2)^3}dxdydz$$ where $$G =\Big\{x^2 + y^2 \le a^2, y^2 + z^2 \le a^2, x \ge 0, y \ge 0, z \ge 0 \Big\}.$$ I suggest that spherical coordinates can help, but I don't know how to compute it in overall

Answer 1

Let $$I:=\iiint_{G_a} \frac{xyz}{(a^2 + x^2 + y^2 + z^2)^3}dxdydz.$$ Cartesian coordinates are fine. $G$ is the intersection of two cylinders in the first octant: $$G_a =\Big\{0\leq y \le \min\left(\sqrt{a^2-x^2},\sqrt{a^2-z^2}\right), x \in[0,a], z\in[0,a] \Big\}.$$ Note that the integral is independent with respect to $a$: letting $X=x/a$, $Y=y/a$ and $Z=z/a$ then $$I=\iiint_{G_1} \frac{a^3 XYZ}{a^6(1 + X^2 + Y^2 + Z^2)^3}(a^3dXdYdZ)=\iiint_{G_1} \frac{XYZ}{(1 + X^2 + Y^2 + Z^2)^3}dXdYdZ$$ Therefore we may assume that $a=1$.

By symmetry of the integrand with respect to the swapping $z\leftrightarrow x$, we have that \begin{align}I &= 2\int_{x=0}^1\int_{z=0}^x\int_{y=0}^{\sqrt{1-x^2}} \frac{xyz}{(1 + x^2 + y^2 + z^2)^3}dydzdx\\ &= -\frac{1}{2}\int_{x=0}^1\int_{z=0}^x\left[\frac{xz}{(1 + x^2 + y^2 + z^2)^2}\right]_{y=0}^{\sqrt{1-x^2}} dzdx\\ &= -\frac{1}{2}\int_{x=0}^1\int_{z=0}^x\frac{xz}{(2 + z^2)^2} dzdx +\frac{1}{2}\int_{x=0}^1\int_{z=0}^x\frac{xz}{(1 + x^2+ z^2)^2} dzdx\\ &= \frac{1}{4}\int_{x=0}^1\left[\frac{x}{2 + z^2} \right]_{z=0}^x dx -\frac{1}{4}\int_{x=0}^1\left[\frac{x}{1 +x^2+ z^2} \right]_{z=0}^x dx\\ &= \frac{1}{4}\int_{x=0}^1\left(\frac{x}{2 + x^2} -\frac{x}{2} -\frac{x}{1 + 2x^2} +\frac{x}{1 + x^2}\right)dx\\ &= \frac{1}{16}\left[2\ln(2+x^2)-x^2-\ln(1+2x^2)+2\ln(1+x^2)\right]_{x=0}^1\\ &=\frac{\ln(3)-1}{16}. \end{align}