hard triple integral

I suspect converting to polars is the easiest way, because otherwise you get a VERY long and tedious integral...

Let $y = x\sinh{t}$ so that $dy = x\cosh{t}\,dt$ and the first integral becomes:

$\int{\frac{(x^2 + y^2)^{\frac{3}{2}}}{16}\,dy} = \int{\frac{[x^2 + (x\sinh{t})^2]^{\frac{3}{2}}}{16}\,x\cosh{t}\,dt}$

$= \int{\frac{(x^2 + x^2\sinh^2{t})^{\frac{3}{2}}}{16}\,x\cosh{t}\,dt}$

$= \int{\frac{[x^2(1 + \sinh^2{t})]^{\frac{3}{2}}}{16}\,x\cosh{t}\,dt}$

$= \int{\frac{(x^2\cosh^2{t})^{\frac{3}{2}}}{16}\,x\c osh{t}\,dt}$

$= \int{\frac{(x\cosh{t})^3}{16}\,x\cosh{t}\,dt}$

$= \int{\frac{x^4\cosh^4{t}}{16}\,dt}$

$= \frac{x^4}{16}\int{\cosh^4{t}\,dt}$

$= \frac{x^4}{16}\int{(\cosh^2{t})^2\,dt}$

$= \frac{x^4}{16}\int{\left(\frac{1}{2}\cosh{2t} + \frac{1}{2} \right)^2\,dt}$

$= \frac{x^4}{16}\int{\frac{1}{4}\cosh^2{2t} + \frac{1}{2}\cosh{2t} + \frac{1}{4}\,dt}$

$= \frac{x^4}{16}\int{\frac{1}{4}\left(\frac{1}{2}\co sh{4t} + \frac{1}{2}\right) + \frac{1}{2}\cosh{2t} + \frac{1}{4}\,dt}$

$= \frac{x^4}{16}\int{\frac{1}{8}\cosh{4t} + \frac{1}{8} + \frac{1}{2}\cosh{2t} + \frac{1}{4}\,dt}$

$= \frac{x^4}{16}\int{\frac{1}{8}\cosh{4t} + \frac{1}{2}\cosh{2t} + \frac{3}{8}\,dt}$

$= \frac{x^4}{16}\left(\frac{1}{32}\sinh{4t} + \frac{1}{4}\sinh{2t} + \frac{3}{8}t\right) + C$

$= \frac{x^4}{16}\left(\frac{1}{16}\cosh{2t}\sinh{2t} + \frac{1}{2}\cosh{t}\sinh{t} + \frac{3}{8}t\right) + C$

$= \frac{x^4}{16}\left[\frac{1}{8}(2\sinh^2{t} + 1)\cosh{t}\sinh{t}+ \frac{1}{2}\sinh{t}\sqrt{1 + \sinh^2{t}} + \frac{3}{8}t\right] + C$

$= \frac{x^4}{16}\left[\frac{1}{8}(2\sinh^2{t} + 1)\sinh{t}\sqrt{1 + \sinh^2{t}} + \frac{1}{2}\sinh{t}\sqrt{1 + \sinh^2{t}} + \frac{3}{8}t\right] + C$

$= \frac{x^4}{16}\left\{\frac{1}{8}\left[2\left(\frac{y}{x}\right)^2 + 1\right]\left(\frac{y}{x}\right)\sqrt{1 + \left(\frac{y}{x}\right)^2} + \frac{1}{2}\left(\frac{y}{x}\right)\sqrt{1 + \left(\frac{y}{x}\right)^2} + \frac{3}{8}\sinh^{-1}\left(\frac{y}{x}\right)\right\} + C$