How to integrate these: $\iiiint_A\frac{t^2dtdxdydz}{t^2+2x^2+3y^2+4z^2}$ and another one in wolfram alpha?

Question

I tried different methods but I cannot put following integrals into Wolfram Alpha in order to find ou results. $$ \iiiint_A\frac{t^2\,dt\,dx\,dy\,dz}{t^2+2x^2+3y^2+4z^2}, $$ where $A=\{(t,x,y,z): t^2+2x^2+3y^2+4z^2<1\}$ and $$\iiiint r^5\sin^2(\alpha_3)\cos(\alpha_2)\cos^2(\alpha_3)\,dr\,d\alpha_1\,d\alpha_2\,d\alpha_3,$$ where $r\in(0,1), ~\alpha_1\in(0,2\pi); ~\alpha_2,\alpha_3\in(-\pi/2,\pi/2)$.

Thank you for help.

Answer 1

I don\t know if it works in Wolfram Alpha, but in Mathematica the second integral can be calculated by

Integrate[ r^5 Sin^2 (x) Cos (y) Cos^2 (z), {r, 0, 1}, {x, 0, 2 Pi}, {y, -Pi/2, Pi/2}, {z, -Pi/2, Pi/2}]

(But is also easy to see that this integral vanishes)

For the first integral, I dont know how to do it directly, but if you use coordinate transformations $x\to \frac 1 {\sqrt{2}} x$, $y \to \frac {1} {\sqrt {3}} y$, $z \to \frac 1 {\sqrt4} z$, then $A\to B_1(0)$ and you can use polar coordinates for $\mathbb{R}^4$ and use the same syntax as for the second example.

Answer 2

Through the substitutions $t=a,\; x=\frac{1}{\sqrt{2}}b,\; y=\frac{1}{\sqrt{3}}c,\; z=\frac{1}{2}d $
and symmetry the given integral boils down to: $$ \frac{1}{2\sqrt{6}}\iiiint_{a^2+b^2+c^2+d^2<1}\frac{a^2}{a^2+b^2+c^2+d^2}\,d\mu\\ = \frac{1}{8\sqrt{6}}\iiiint_{a^2+b^2+c^2+d^2<1}\frac{a^2+b^2+c^2+d^2}{a^2+b^2+c^2+d^2}\,d\mu$$ and that is just $\frac{1}{8\sqrt{6}}$ times the (hyper-)volume of the unit sphere in $\mathbb{R}^4$, i.e. $\large\color{red}{\frac{\pi^2}{16\sqrt{6}}}$.