Calculate a triple integral over a region.

Question

Calculate integral $\iiint\limits_Gf(x;y;z)dxdydz$ if $$ f=\sqrt{x^2+y^2},\ G=\{x^2+y^2+z^2\leqslant31,z\leqslant1\} $$

At first I tried to use Spherical coordinates change of variables, but the attempt was rather unsuccessful since the inequality $z\leqslant1$ complicated the solution very much.
So, then I tried to use Cylindrical coordinates change, but it didn't really work the way I wanted, and here is why: $$ \begin{aligned} &\begin{cases} x=r\cos\phi,\\ y=r\sin\phi,\\ z=z \end{cases}\Rightarrow \begin{cases} f=r,\\ x^2+y^2+z^2=r^2+z^2\leqslant31,\\ z\leqslant1\text{ (stays the same)} \end{cases}\\ &\text{Besides, } |J|=r \end{aligned} $$ $$ \begin{aligned} &\iiint\limits_Gf(x;y;z)dxdydz=\int\limits_{-\sqrt{31}}^1dz\int\limits_{-\sqrt{31-z^2}}^0r^2dr+\int\limits_{-\sqrt{31}}^1dz\int\limits_0^{\sqrt{31-z^2}}r^2dr=\dots=\\ &=\frac{2}{3}\int\limits_{-\sqrt{31}}^1(31-z^2)^\frac{3}{2}dz \end{aligned} $$ And that last integral unsettles me because its calculation is quite difficult. So, I thought that there might be a better solution to this problem, excluding the necessity of calculating difficult integrals.
However, I don't deny that I might have mistaken. So, if someone could help in any way, I would be appreciative.

Answer 1

In cylindrical coordinates, your integral becomes$$\int_0^{2\pi}\int_0^{\sqrt{31}}\int_{-\sqrt{31-r^2}}^1r^2\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=2\pi\int_0^{\sqrt{31}}r^2\left(1+\sqrt{31-r^2}\right)\,\mathrm dr.\tag1$$Now, using the fact that$$\int r^2\sqrt{a-r^2}\,\mathrm dr=\frac18\left(a^2\arctan\left(\frac r{\sqrt{a-r^2}}\right)+r\sqrt{a-r^2}\left(2 r^2-a\right)\right),$$you get that $(1)$ is equal to $\frac{31}{24} \pi \left(16 \sqrt{31}+93 \pi \right)$.

Answer 2

Hint. For the last integral, if you let $z=\sqrt{31}\sin\phi,$ then you only have to integrate a constant multiple of $$\cos^4\phi\mathrm d\phi,$$ which you may do relatively easily if you recall that $$\cos^4\phi-\sin^4\phi=(\cos^2\phi-\sin^2\phi)(\cos^2\phi+\sin^2\phi)=\cos2\phi,$$ and that $$2(\cos^4\phi+\sin^4\phi)=(\cos^2\phi+\sin^2\phi)^2+(\cos^2\phi-\sin^2\phi)^2=1+\cos^22\phi,$$ so that $$4(\cos^4\phi+\sin^4\phi)=2+1+\cos4\phi=3+\cos4\phi.$$

Or better still, just recall that $$\cos^4\phi=\frac14(2\cos^2\phi)^2=\frac14(1+\cos2\phi)^2=\frac14(1+2\cos2\phi+\cos^22\phi),$$ etc., since the last summand is easily linearised by $2\cos^22\phi=1+\cos4\phi.$