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The spherical coordinates of a point can be obtained from its Cartesian coordinates $(x, y, z)$ by the formulae

${\displaystyle {\begin{aligned}r&={\sqrt {x^{2}+y^{2}+z^{2}}}\\\theta &=\arccos {\frac {z}{\sqrt {x^{2}+y^{2}+z^{2}}}}=\arccos {\frac {z}{r}}\\\varphi &=\arctan {\frac {y}{x}}\end{aligned}}}$

The Cartesian coordinates may be retrieved from the spherical coordinates by

${\displaystyle {\begin{aligned}x&=r\,\sin \theta \,\cos \varphi \\y&=r\,\sin \theta \,\sin \varphi \\z&=r\,\cos \theta \end{aligned}}} $

A function $f(r,\theta,\varphi)$ can be integrated over every point in $\mathbb{R}^3$ by the triple integral

${\displaystyle \ \int \limits _{\varphi =0}^{2\pi }\ \int \limits _{\theta =0}^{\pi }\ \int \limits _{r=0}^{\infty }f(r,\theta ,\varphi )r^{2}\sin \theta \,\mathrm {d} r\,\mathrm {d} \theta \,\mathrm {d} \varphi .}$

If we have a triple integral of the form

$\displaystyle \int _0^{\infty }\int _0^{\infty }\int _0^{\infty }f(x,y,z)dxdydz$

the corresponding integral in spherical coordinates will be

$\displaystyle \int_{\varphi=\gamma}^{\varphi=\psi} \int_{\theta=\alpha}^{\theta=\beta} \int_{r=a}^{r=b} f(r,\theta,\varphi) \, r^2 \sin \, \theta \, dr \, d\theta \, d\varphi.$

For the bounds of integration of $x,y,z$ going from zero to infinity, what would be the values of $\gamma,\psi,\alpha,\beta,a,b$ ?

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    $\begingroup$ didn't you just mention the answer? $\endgroup$ – MegaX 824 Jun 8 at 4:40
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${\displaystyle \ \int \limits _{\varphi =0}^{\pi/2 }\ \int \limits _{\theta =0}^{\pi/2 }\ \int \limits _{r=0}^{\infty }f(r,\theta ,\varphi )r^{2}\sin \theta \,\mathrm {d} r\,\mathrm {d} \theta \,\mathrm {d} \varphi .}$

We limit $\varphi$ from $0$ to $\frac{π}{2}$ to keep $z$ positive, and we also keep $\theta$ from $0$ to $\frac{π}{2}$ so that both $x$ and $y$ are both positive.

Then we would have $\gamma=0,\psi=\pi/2,\alpha=0,\beta=\pi/2,a=0,b=\infty$.

Although equivalent, the convention I've learned is that $\phi\in(0,\pi)$ and $\theta\in(0,2\pi)$

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As the region over which you are integrating is the 1st octant of $\mathbb R^3$, there are only bounds on $\theta$ and $\phi$. $r$ is still unbounded i.e. $a=0$, and $b=\infty$

Now, as $\theta$ is the angle made with the $z$ axis, as we are integrating only over half of the $z$ axis, the range of $\theta$ is also halved. Also, as the half is positive, we have $\alpha = 0$ and $\beta = \frac{\pi}{2}$.

Similarly, as $\phi$ goes over 1 quarter of the x-y plane (the 1st quadrant), we have $\gamma = 0$, and $\psi = \frac{\pi}{2}$

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