Quadruple integral $\iiiint_E \sin(x^2+y^2) \,dx\,dy\,dz\,dt$ Calculate the integral over E $$\iiiint_E \sin(x^2+y^2) \,dx\,dy\,dz\,dt$$ where E is
$$\{ z^2 \leqslant x^2+y^2 \leqslant 2z;\\ t \leqslant xy \leqslant 2t\}$$

To identify the integration limits I came up with the idea to apply the inequality $$xy \leqslant \frac{x^2 + y^2}{2}$$ and therefore claim that $$t \leqslant z$$ Also $$t \leqslant 2t$$ and $$z^2 \leqslant 2z$$ imply that $$t \geqslant 0$$ and $$0 \leqslant z \leqslant 2 $$ The last step I did was making a substitution $$x=r\cos\phi;\ y=r\sin\phi$$ from which I got $$t \leqslant r^2\cos\phi \sin\phi \leqslant 2t$$ and (using the fact that $\sin(x^2+y^2)$ is even, so to obtain the integral from $-\pi$ to $\pi$ we can double the integral from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$)  $$ \frac{\arcsin\frac{2t}{r^2}}{2} \leqslant \phi \leqslant \frac{\arcsin\frac{4t}{r^2}}{2} $$
As a result, I got the following repeated integral:
$$ 2\int_{0}^{2}\,dz\int_{0}^{z}\,dt\int_{z}^{\sqrt{2z}}\,dr\int_{\frac{\arcsin\frac{2t}{r^2}}{2}}^{\frac{\arcsin\frac{4t}{r^2}}{2}}r\sin{r^2}\,d\phi $$
But trying to calculate this, I get the integral of $\arcsin$ difference which I cannot take. Are there different ways to get a repeated integral from the given quadruple one?
 A: Assuming we can do the $xy$ integrals first, we have two simultaneous inequalities that need to be satisfied due to AM-GM:
$$\begin{cases}4t \leq 2z \\ 2t \leq z^2  \end{cases}$$
since the upper bounds for the $t$ always need to be less than the upper bounds for the $z$, and same for the lower bounds. So we have a situation where we are sometimes limited by our upper bounds and sometimes by the lower bounds. This means we have to split up the outer integrals like so:
$$\iint_{tz} = \int_0^1 \int_0^{\frac{z^2}{2}} + \int_1^2\int_0^{\frac{z}{2}}$$
For the inner integrals, let's do polar coordinates as you suggest and take note of the same $x \leftrightarrow y$ exchange symmetry since our integral is only in the first quadrant. This will also be split into three integrals, giving a total of six integrals. 
$$\frac{I}{2} = \int_0^1 \int_0^{\frac{z^2}{2}} \int_{\frac{1}{2}\sin^{-1}\left(\frac{t}{z}\right)}^{\frac{1}{2}\sin^{-1}\left(\frac{2t}{z}\right)} \int_{\sqrt{\frac{2t}{z^2\sin 2\phi}}}^{2z} + \int_0^1 \int_0^{\frac{z^2}{2}} \int_{\frac{1}{2}\sin^{-1}\left(\frac{2t}{z}\right)}^{\frac{1}{2}\sin^{-1}\left(\frac{2t}{z^2}\right)}  \int_{\sqrt{\frac{2t}{z^2\sin 2\phi}}}^{\sqrt{\frac{2t}{z\sin 2\phi}}}$$ $$ + \int_0^1 \int_0^{\frac{z^2}{2}} \int_{\frac{1}{2}\sin^{-1}\left(\frac{2t}{z}\right)}^{\frac{\pi}{4}}  \int_{z^2}^{\sqrt{\frac{2t}{z\sin 2\phi}}} + \cdots$$
(with the other three integrals being similar, but swapping the middle two angles) and with an integrand of
$$r\sin r^2$$ This can also be done with just four integrals by doing the angular integral first (as it would split the integral into two regions), but this leads to the same problem of integrating the product of a trig and inverse trig. In short, I don't see a way to swap the integrals further.
