Shifted spherical coordinates

Question

I need to solve this integral: $\int _V\sqrt{x^2+y^2+z^2}dx\:dy\:dz$ where V is described with $x^2+y^2+z^2\le x$.

So, I know I could use shifted spherical coordinates but then the integral is hard to solve because of the function. But if I use normal spherical coordinates ( where the origin is in (0,0,0) ) it would be much easier to solve, but it's harder to find the boundaries for radius R. Does it depend both on θ and ϕ, or just on one of them?

(I'm using standard notation for angles, see here:https://en.wikipedia.org/wiki/Spherical_coordinate_system).

Could anyone be kind to help me find boundaries for R?

Thank you!

Answer 1

If you use spherical conversions, you get the equation

$$0 \leq \rho^2 \leq \rho \sin(\phi) \cos(\theta) \implies 0 \leq \rho \leq \sin(\phi)\cos(\theta)$$

So long as you are careful about how you do your bounds for $\phi$ and $\theta$ wouldn't that solve your problem?

Answer 2

Changing into spherical coordinates $$ \left\{\begin{aligned}x&=r\,\sin \theta \,\cos \varphi \\y&=r\,\sin \theta \,\sin \varphi \\z&=r\,\cos \theta \end{aligned}\right. \quad r\ge 0, \quad 0\le \theta \le \pi, \quad 0\le \varphi \le 2\pi $$ the domain of integration becomes $$ \tilde{V}=\{(r,\theta,\varphi)|\quad 0\le r\le \sin\theta\cos\varphi,\quad 0\le \theta \le \pi, \quad 0\le \varphi \le 2\pi\}. $$ Since \begin{eqnarray} \dfrac{\partial(x,y,z)}{\partial(r,\theta,\varphi)}&=&\left|\begin{array}{ccc}\sin\theta\cos\varphi&r\cos\theta\cos\varphi&-r\sin\theta\sin\varphi\\ \sin\theta\sin\varphi&r\cos\theta\sin\varphi&r\sin\theta\cos\varphi\\ \cos\theta&-r\sin\theta&0\end{array}\right|\\ &=&\cos\theta\left|\begin{array}{cc}r\cos\theta\cos\varphi&-r\sin\theta\sin\varphi\\ r\cos\theta\sin\varphi&r\sin\theta\cos\varphi\end{array}\right|+r\sin\theta\left|\begin{array}{cc}\sin\theta\cos\varphi&-r\sin\theta\sin\varphi\\ \sin\theta\sin\varphi&r\sin\theta\cos\varphi\end{array}\right|\\ &=&r^2\cos^2\theta\sin\theta+r^2\sin^3\theta=r^2\sin\theta \end{eqnarray} we have \begin{eqnarray} \int_V\sqrt{x^2+y^2+z^2}\,dx\,dy\,dz&=&\int_{\tilde{V}}r(r^2\sin\theta)\,dr\,d\theta\,d\varphi=\int_0^{2\pi}\,d\varphi\int_0^{\pi}\,d\theta\int_0^{\sin\theta\cos\varphi}r^3\sin\theta\,dr\\ &=&\dfrac14\int_0^{2\pi}\cos^4\varphi\,d\varphi\int_0^{\pi}\sin^5\theta\,d\theta=\dfrac12A_4B_5 \end{eqnarray} with \begin{eqnarray} A_4&=&\int_0^\pi\cos^4\varphi\,d\varphi=\int_{-\pi/2}^{\pi/2}\sin^4t\,dt=2\int_0^{\pi/2}\sin^4\,dt=2W_4=\dfrac{3\pi}{8}\\ B_5&=&\int_0^\pi\sin^5\theta\,d\theta=\int_{-\pi/2}^{\pi/2}\cos^5t\,dt=2\int_0^{\pi/2}\cos^5t\,dt=2W_5=\dfrac{16}{15} \end{eqnarray} Hence $$ \int_V\sqrt{x^2+y^2+z^2}\,dx\,dy\,dz=\dfrac12A_4B_5=\dfrac12\cdot\dfrac{3\pi}{8}\cdot\dfrac{16}{15}=\dfrac\pi5. $$