Can the following triple integral be computed via elementary calculus methods?

Solution 1:

$$ \begin{align} &\int_0^{2\pi}\int_0^1\int_0^1xy\sqrt{x^2+y^2-2xy\cos(\theta)}\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}\theta\tag{1}\\ &=2\int_0^{2\pi}\int_0^1\int_0^yxy\sqrt{x^2+y^2-2xy\cos(\theta)}\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}\theta\tag{2}\\ &=2\int_0^{2\pi}\int_0^1\int_0^1ry^2\sqrt{r^2y^2+y^2-2ry^2\cos(\theta)}\,\,y\,\mathrm{d}r\,\mathrm{d}y\,\mathrm{d}\theta\tag{3}\\ &=2\int_0^{2\pi}\int_0^1\int_0^1ry^4\sqrt{r^2+1-2r\cos(\theta)}\,\mathrm{d}y\,\mathrm{d}r\,\mathrm{d}\theta\tag{4}\\ &=\frac25\int_0^{2\pi}\int_0^1\sqrt{r^2+1-2r\cos(\theta)}\,r\,\mathrm{d}r\,\mathrm{d}\theta\tag{5}\\ &=\frac25\int_{-\pi/2}^{\pi/2}\int_0^{2\cos(\theta)}r^2\,\mathrm{d}r\,\mathrm{d}\theta\tag{6}\\ &=\frac2{15}\int_{-\pi/2}^{\pi/2}8\cos^3(\theta)\,\mathrm{d}\theta\tag{7}\\ &=\frac{16}{15}\int_{-1}^1(1-u^2)\,\mathrm{d}u\tag{8}\\[4pt] &=\frac{64}{45}\tag{9} \end{align} $$ Explanation:
$(2)$: the integral is the same for $x\lt y$ as for $x\gt y$, so assume $x\lt y$ and multiply by $2$
$(3)$: substitute $r=\frac xy$
$(4)$: collect the $y$s and switch the order of integration
$(5)$: integrate in $y$
$(6)$: $(5)$ is the distance from $(1,0)$ integrated over the unit disk centered at $(0,0)$;
$\hphantom{(6):}$this is the same as $r$ integrated over the unit disk centered at $(1,0)$,
$\hphantom{(6):}$whose equation is $r\le2\cos(\theta)$ for $\theta\in[-\pi/2,\pi/2]$
$(7)$: integrate in $r$
$(8)$: substitute $u=\sin(\theta)$
$(9)$: integrate in $u$