# Convert a triple integral into an iterated one.

I have convert a triple integral $$\left(\iiint\limits_Gf(x;y;z)dxdydz\right)$$ into an iterated one $$\left(\int\limits_?^?d\phi\int\limits_?^?d\psi\int\limits_?^?fr^2\cos\psi dr \right)$$ where $$f$$ is not given and $$G=\{x^2+y^2+z^2\leqslant2az,\ x^2+y^2\geqslant z^2\}$$

I did the following (using Spherical coordinates): \begin{aligned} &1)\ r^2\leqslant2ar\sin\psi\Rightarrow 0\leqslant r\leqslant2a\sin\psi\\ &2)\ x^2+y^2=r^2\cos^2\psi(\cos^2\phi+\sin^2\phi)=r^2\cos^2\psi\geqslant r^2\sin^2\psi=z^2\Rightarrow\\ &\Rightarrow \cos^2\psi-\sin^2\psi=\cos2\psi\geqslant0\Rightarrow -\frac{\pi}{4}\leqslant\psi\leqslant\frac{\pi}{4}\\ &3)\ x^2+y^2+z^2\leqslant2az\iff x^2+y^2+(z-a)^2\leqslant a^2\Rightarrow \psi\geqslant 0\Rightarrow0\leqslant\psi\leqslant\frac{\pi}{4}\\ &4)\ 0\leqslant\phi\leqslant2\pi \end{aligned} But I am not sure about $$\psi$$ and especially $$\phi$$.
So, if someone could check my solution, I would appreciate it.

Note that the region G is enclosed between a sphere of center $$(0,0,a)$$ and radius $$a$$ , and a cone of 45-degree angle and with its vertex at origin.
In spherical coordinates, the region is bounded by the sphere given by $$r=2a\cos\phi$$ and $$0\le \phi \le \frac\pi4$$. Thus, the integral reads,
$$\iiint\limits_Gf(x,y,z)dxdydz =\int_0^{2\pi}d\theta\int_0^{\pi/4}\sin\phi d\phi\int_0^{2a\cos\phi}r^2f(r,\phi,\theta)dr$$
• Thank you! It turned out that my solution is the same as yours, but I used slightly different coordinate substitution: $x=r\cos\phi\cos\psi,\ y=r\sin\phi\cos\psi,\ z=r\sin\psi,\ |J|=r^2\cos\psi$. Dec 12, 2019 at 15:06