# Bounds of integration in spherical coordinates

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}}}

{\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 $$\gamma,\psi,\alpha,\beta,a,b$$ ?

• didn't you just mention the answer? – MegaX 824 Jun 8 at 4:40

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

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 $$\theta\in(0,2\pi)$$

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 $$\psi = \frac{\pi}{2}$$