# Using spherical coordinates find the limits of integration of the region inside a sphere with center $(a,0,0)$ and radius $a$

I am asked to find the limits of integration (using spherical coordinates) of the region inside a sphere with center $(a,0,0)$ and radius $a$.

I got to this point:

\begin{align*} (x-a)^2+y^2+z^2 &= a^2\\ x^2-2ax+a^2+y^2+z^2 &= a^2\\ \rho^2 - 2a \rho \cos(\theta) \sin(\phi) &= 0\\ \rho &= 2 a \cos(\theta) \sin(\phi) \end{align*}

Knowing that $0 \leq \phi \leq pi$ and $- \frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$ one can write

$$\int_{- \frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{\pi} \int_{0}^{2 a \cos(\theta) \sin(\phi)} f(\rho \cos(\theta) \sin(\phi),\rho \sin(\theta) \sin(\phi), \rho cos(\phi)) \ \rho^2 \sin(\phi) \ d\rho d\phi d\theta$$

Is my reasoning correct?

Thank you.

• Let's say $$a \neq 0$$, otherwise we wouldn't have a sphere. – bru1987 May 9 '17 at 12:57

## 1 Answer

Looks good to me!

Note that shifting the origin-centered sphere to $(a,0,0)$ moves the sphere completely to the octants where $x \ge 0$. The projection onto the $xy$-plane is a circle of radius $a$ centered at $(a,0)$ and from there, it's easy to see $\theta$ runs from $-\tfrac{\pi}{2}$ to $\tfrac{\pi}{2}$.

For $\phi$, note that the projection of the sphere onto the $yz$-plane is still a circle with center in the origin (and radius $a$) so from there, it's clear that $\phi$ still runs from $0$ to $\pi$.

For $\rho$, you correctly obtained the upper bound from there sphere's equation by substitution of the spherical coordinates so you indeed arrive at: $$\int_{- \frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{\pi} \int_{0}^{2 a \cos(\theta) \sin(\phi)} f(\rho \cos(\theta) \sin(\phi),\rho \sin(\theta) \sin(\phi), \rho cos(\phi)) \ \rho^2 \sin(\phi) \ d\rho d\phi d\theta$$ Note that taking $f(x,y,z)=1$ leads to: $$\int_{- \frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{\pi} \int_{0}^{2 a \cos(\theta) \sin(\phi)} \rho^2 \sin(\phi) \ d\rho d\phi d\theta = \ldots = \frac{4}{3}\pi a^3$$ which is indeed the volume of a sphere with radius $a$.

• Thank you my friend, have a great day! – bru1987 May 9 '17 at 14:37
• @bru1987 You're welcome; thanks and the same to you! – StackTD May 9 '17 at 14:37