# Visualizing solids of revolution

Can someone shed a little bit of light on the problem of volumes of revolution about the $x$-axis and the $y$-axis of the same shapes? Take for example $f(x)=x^2$. If we want to find the volume bounded by the parabola and the $x$-axis with axis of revolution at $y=0$ we would use the standard method of disks and get the volume $\pi/5$ for $x=[0,1]$. Now if we want to find the volume by rotating the curve around the $y$-axis instead, using the cylindrical shells method, we get the volume to be $\pi/2$.

Now, maybe it's just my flawed intuition, but since we are basically rotating the same shape/area by 360$^\circ$ (but in different "directions"), I would guess the two volumes to be the same. Anyone knows some good graphics/animation/resources to help me visualize this problem?

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Note also that for rotating about y, you could delete the whole area to the left of the axis and not change the volume.  Rotating about x you cannot. – Ross Millikan May 16 '12 at 15:36
@RossMillikan the area to the left that you speak of, is as if it was "deleted" when integrating for $x=[0,1]$, isn't it? Or am I missing your point? – Milosz Wielondek May 16 '12 at 15:45
my point was that it is another way of seeing the disconnect between area and rotated volume.  You could have it or not when rotating about y and get the same volume, but the area changes by a factor 2. – Ross Millikan May 16 '12 at 16:00

## 1 Answer

When you rotate about $y$, most of the area is farther from the axis than if you rotate around $x$. If you think about a small area of size $dxdy$ when rotated around $x$ it sweeps out a volume element $ydxdy$ and when rotated around $y$ is sweeps out a volume $xdxdy$. The methods of discs and shells just do one dimension of the integral for you.

You could think of a rectangle $[0,10] \times [0,0.1]$. If you rotate it around $x$ you get a cylinder of radius $0.1$ and height $10$, for volume $0.1\pi$. If you rotate it around $y$ the radius is $10$ and the height is $0.1$ for a volume of $10\pi$

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Naturally, it does make perfect sense mathematically. But am I the only one to think it's somewhat counter-intuitive? – Milosz Wielondek May 16 '12 at 16:25
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