# Find volume of a revolved solid by integrating wedges.

So, lets say that I wanted to find the volume of the solid formed by rotating the area between
$f(x)=\sqrt{1-x^2}, 0<x<1$ and the $x$ axis around the $y$ axis. (This example is simply a hemisphere).

Now normally, I would use geometry, or the "disk method", so the area would simply be $\pi\int_0^1(1-y^2)dy=\frac{2\pi}{3}$.

I was thinking about this and I was wondering if it would be possible to find the answer by integrating wedges of this volume from $0$ to $2\pi$. This seems to be an approach that more closely resembles the premise of the problem. At first I thought that this might be as easy as $\frac{1}{2}\int_0^{2\pi}[\int_0^1f(x)dx]^2d\theta$, essentially integrating a polar circle with radius of the area that is revolved around the y axis. However, when I tried this, I did not get my expected answer. I calculated the volume to be $\frac{\pi^3}{16}$, however, I should have found the volume to be $\frac{2\pi}{3}$.

Can anyone help me understand why my approach was not successful, and also explain a successful method of evaluating the volume in this way?

• The problem is with the way you have choosen your differential element. Since you are finding volume in $\mathbb{R}^3$ using cylindrical co-ordinates(unknowingly I guess.Clarify me) Your Integral should look like $\int{ \int \int dzdrd\theta}$. – Ramana Venkata May 16 '12 at 5:00
• @RamanaVenkata I am using Cartesian coordinates. joriki's answers seems to work perfectly, but thank you for your help anyway. – diracdeltafunk May 16 '12 at 5:30
• Indirectly your using cylindrical coordinates. Even joriki's answer also written using cylindrical coordinates in some sense. – Ramana Venkata May 16 '12 at 5:37
• @RamanaVenkata I suppose that's true, however I definitely feel more comfortable in Cartesian or polar. – diracdeltafunk May 16 '12 at 5:43

In a wedge with angular extent $\mathrm d\theta$, an area $\mathrm dS$ of the rotated quarter-circle contributes $x\mathrm dS\mathrm d\theta$ to the volume of the wedge, so the volume is
\begin{align} \int_0^{2\pi}\left[\int x\mathrm dS\right]\mathrm d\theta &=\int_0^{2\pi}\left[\int_0^1xf(x)\mathrm dx\right]\mathrm d\theta \\ &=\int_0^{2\pi}\left[\int_0^1x\sqrt{1-x^2}\mathrm dx\right]\mathrm d\theta \\ &=\int_0^{2\pi}\left[-\frac13\sqrt{1-x^2}^3\mathrm dx\right]_0^1\mathrm d\theta \\ &=\frac{2\pi}3 \end{align}
• @Ben: Sorry, I don't understand the reasoning in your second comment; you'll have to elaborate. In response to your first comment, I didn't say that a wedge has volume $x\mathrm dS\mathrm d\theta$, but that an area $\mathrm dS$ of the rotated quarter-circle contributes $x\mathrm dS\mathrm d\theta$ to the volume of a wedge with angular extent $\mathrm d\theta$. This is because the part of the wedge corresponding to a surface element $\mathrm dS$ of the rotated quarter-circle is to first order a prism with base area $\mathrm dS$ and height $x\mathrm d\theta$. – joriki May 16 '12 at 6:14
• I think your terminology is confusing me a bit. First, why are you calling the area $dS$? Why not $S$? Also, how is the volume of the prism equal to $xdSd\theta$? Can you offer a proof of this? – diracdeltafunk May 17 '12 at 2:57