# Volume of a sphere (solid of revolution)

1. Jan 25, 2009

Hi guys, I did post this over on the h/w c/w forum last week, but I think actually it is out of the scope of that, plus its not actually h/w c/w, its my general wondering :D

So I am having an issue with the volume of a sphere and calculating it. I am using the method where the solid, sphere in this case, is composed of many cylinders, in my case along the x-axis where the center of the sphere is at the origin.

So initially I attempted this by considering each cylinder was of volume:$\pi y^2 \delta x$
then the formula of the circle in the xy plane of the sphere is $x^2 + y^2 = r^2$ which yields $y^2 = {r^2 - x^2}$ which means the volume of each cylinder becomes $\pi (r^2 - x^2) \delta x$

I will limit my range to [0,r] as it is symmetrical about the yz plane. So integrating accros the interval [0,r] we get :

$$V = \pi \int_{0}^{r} (r^2 - x^2) \delta x$$
$$= \pi ({r^2}x - \frac{x^3}{3})\bigg{|}_0^r$$
$$= \pi [({r^3} - \frac{r^3}{3})-({r^3}(0) - \frac{0^3}{3})]$$
$$= \pi \frac{2r^3}{3}$$

then x2 to get the volume of a sphere to be ${4\pi r^3}{/}{3}$

now thats all well and good, but I also wanted to see if another method would work, instead of using x as our independent use theta. Now I learnt that if you wanted to calculate the surface are of a sphere you can use theta as it makes calculations easier. So using the idea that we can calculate the surface are of a sphere modeling it as composed of many hollow and cylinders with no base or head, each has height $r\delta \theta$. So each cylindrical surface becomes :

$$i. \ A = 2\pi y r \delta \theta$$
$$ii. \ y=rsin\theta$$
$$iii. \ A = 2r\pi rsin\theta \delta \theta$$

So thats fine, that idea can be taken as can successfully yield the surface area of a sphere successfully. Now I figured if that method can be employed for a surface, it must be adaptable to find the volume of a sphere. so this time I decided to define each solid cylinder as:

$$iv. \ V = \pi y^2 r \delta \theta$$
$$v. \ V = r\pi (r^2sin^2\theta) \delta \theta$$

so again I approached this as before, this time integrating across the range [0, $\pi / 2$] where I again will only find the hemisphere for +ve x and theta is the angle from the x axis. So my integration went like this :

$$V = \pi r^3\int_{0}^{\frac{\pi}{2}} (sin^2\theta) \delta \theta$$
$$= \pi r^3\int_{0}^{\frac{\pi}{2}} \frac{1}{2}(1 - cos2\theta) \delta \theta$$
$$= \frac{\pi r^3}{2}(\theta - \frac{1}{2}sin2\theta)\bigg{|}_{0}^{\frac{\pi}{2}}$$
$$= \frac{\pi r^3}{2}(\frac{\pi}{2}) = \frac{\pi^2 r^3}{4}$$

then oviously x2, but as is evident it is not the right answer, I really cant see what I have done wrong, I pretty sure none of my actually mathematics is wrong, which means the only part that could be wrong is the $r\delta \theta$ bit and teacher suggested that for some reason it doesn't apply to a solid but does to a shell, but couldn't give a valid reason as to why. I am inclined to agree with him, but again I cant see why. In various texts I have looked though it says things along the lines of for the shell it is tempting to use $\delta x$ but that $r\delta \theta$ is a "better approximation", therefore surely it to is a better approximation for a solid of revolution to.

Can anyone explain why this is, what is the reason the integration cant be done this way, or am have I simply done my maths wrong, thanks guys :D

2. Jan 25, 2009

### CompuChip

If I let $\theta$ be the azimuthal component then I get for the height of the cylinder $r \cos\theta$ with $\theta$ running from 0 to pi.
Also you need to include the Jacobian from the coordinate transformation into polar coordinates, which I think (since r is fixed) is just sin(theta) - you should check this.
Then I find
$$\pi r^3 \int_{0}^{\pi} \cos^2\theta (\sin\theta \delta \theta) = \frac{2 \pi r^3}{3}$$
for half the volume.

I hope that gets you started.