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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?

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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. –  Ben7005 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. –  Ben7005 May 16 '12 at 5:43

1 Answer 1

up vote 3 down vote accepted

You can certainly integrate using wedges instead of disks, but what you've written down doesn't correspond to wedges. One thing you can check to see that this can't be right is dimensions – your expression has dimensions of length to the fourth power whereas a volume has length to the third. That's not entirely suprising when you say "with radius of the area" – it doesn't make sense to use an area as a radius (at least not usually :-).

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

$$ \begin{align} \int_0^{2\pi}\left[\int x\mathrm dS\right]\mathrm d\theta &=\int_0^{2\pi}\left[\int_0^1xf(x)\mathrm dx\right]\mathrm d\theta \\ &=\int_0^{2\pi}\left[\int_0^1x\sqrt{1-x^2}\mathrm dx\right]\mathrm d\theta \\ &=\int_0^{2\pi}\left[-\frac13\sqrt{1-x^2}^3\mathrm dx\right]_0^1\mathrm d\theta \\ &=\frac{2\pi}3 \end{align} $$

as expected.

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This answer does work, however, I do not understand why the wedge might have a volume of xdSdθ. Cna you clarify this? –  Ben7005 May 16 '12 at 5:27
    
Also, my reasoning behind my method was that if you have a length and find the area of a circle with that method, you "should" be able to take an area and find a volume. –  Ben7005 May 16 '12 at 5:29
    
@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? –  Ben7005 May 17 '12 at 2:57

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