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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.

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1 Answer 1

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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,

$$\iiint\limits_Gf(x,y,z)dxdydz =\int_0^{2\pi}d\theta\int_0^{\pi/4}\sin\phi d\phi\int_0^{2a\cos\phi}r^2f(r,\phi,\theta)dr$$

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  • $\begingroup$ Thank you! It turned out that my solution is the same as yours, but I used slightly different coordinate substitution: $x=r\cos\phi\cos\psi,\ y=r\sin\phi\cos\psi,\ z=r\sin\psi,\ |J|=r^2\cos\psi$. $\endgroup$
    – Bonrey
    Dec 12, 2019 at 15:06

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