# How to solve this problem using spherical coordinates system?

The question is very simple:

Volume inside the solid limited by:$(X^2+Y^2+Z^2=16), (X^2+Y^2=4)$

using SPHERICAL coordinates system.

The final answer however can be checked making a "cylindrical integration".

• The answer is simply the volume of a sphere with radius 4 – Intrepid Traveller Jul 24 '15 at 2:19
• Actually the solid is a regular cylinder with a "piece" of sphere at the top (graph the functions if this is confused). Anyway, the answer for this question is 46.97 for each cylinder. (We have 2: one in the positive "Z" axis, and another cylinder in the negative "Z" axis). The problem is not how to resolve it... is how to resolve using spherical coordinates... – Bruno de Assis Jul 24 '15 at 2:32 In general, when you integrate a volume that includes the origin, $\theta$ and $\phi$ are integrated between 0and $\pi$ and 0 and $2\pi$, respectively. If the volume is convex (or, more generally, star-shaped as it is called) then the shape is determined by the surface function $r(\theta,\phi)$. For example, in the case of a sphere of radius 4 it is $r(\theta,\phi)=4$, in the case of an spheroid is, for example, $r(\theta,\phi)=(1+0.5\cos\theta)$, in the case of a cube is very complicated, so dont use spherical coordinated to calculate the volume of a cube, but in principle there is no problem.

In this case, as you point out, the volume is a cylinder with height 4 and sphere pieces on the top and bottom. N the direction of these pieces,the surface is just the sphere, and in the other directions the surface of the cylinder fullfils $\sqrt{x^2+y^2}/r(\theta,\phi)=\sin\theta$. The equations below just summarizes all this

$\int_0^{2\pi}d\phi\int_{0}^{\pi}\sin\theta d\theta \int_0^{r(\theta)}r^2 d r$ where $r(\theta)= \begin{cases} 4 & \theta<\pi/6~{\rm or}~\theta>5\pi/6\\ 2/\sin\theta & \neg \end{cases}$

We can integrate in the following way

\begin{align} \int_0^{2\pi}d\phi\int_{0}^{\pi}\sin\theta d\theta \int_0^{r(\theta)}r^2 d r\\ &= \int_0^{2\pi}d\phi\int_{0}^{\pi}\sin\theta d\theta \frac{r(\theta)^3}{3}\\ &= 2\pi\int_{0}^{\pi}\sin\theta d\theta \frac{r(\theta)^3}{3}\\ &=2\pi\left(\int_0^{\pi/6}\frac{r(\theta)^3}{3} \sin\theta d\theta +\int_{5\pi/6}^{\pi}\frac{r(\theta)^3}{3} \sin\theta d\theta + \int_{\pi/6}^{5\pi/6}\frac{r(\theta)^3}{3} \sin\theta d\theta\right)\\ &=2\pi\left(\int_0^{\pi/6}\frac{64}{3} \sin\theta d\theta +\int_{5pi/6}^{\pi}\frac{64}{3} \sin\theta d\theta + \int_{\pi/6}^{5\pi/6}\frac{8}{3\sin^3\theta} \sin\theta d\theta\right)\\ &=2\pi \left(64\frac{(2-\sqrt{3})}{3}+\int_{\pi/6}^{5\pi/6}\frac{8\csc^2\theta}{3} d\theta\right)\\ &=2\pi \left(64\frac{(2-\sqrt{3})}{3}+\frac{16}{\sqrt{3}} \right)\\ &\sim 93.96 \end{align} Note that the integrals are symmetric respecto to $\theta=\pi/2$.

• Can you explain to me how you get this? – Bruno de Assis Jul 24 '15 at 3:53
• When I read your answer, I found the following: $\int_0^{2\pi}d\phi\int_{0}^{\pi/2}\sin\theta d\theta \int_{2/sin\theta}^{4}2r^2 d r$ ..I mean, I put a "2" in the integrand and I changed the lower limit for de Last Integral... Is That Correct? Because, doing what I said, you get a Volume of 93.9, wich is the correct answer... – Bruno de Assis Jul 24 '15 at 13:35
• I don't know about your integral. Note that there, sometimes $2/sin(\theta)>4$, so the integral would be negative. I also get 93.9578 when integrating the expression I give in my answer. – Enredanrestos Jul 24 '15 at 14:03
• Yes, I got your point, what you said made sense so I believe that your integral is absolutely correct, but I'm having problem to evaluate it... It is a very elementar problem, so really really sorry, but can you put the steps to evaluate your expression? – Bruno de Assis Jul 24 '15 at 15:59