# Can someone point out what I'm doing wrong here with my solids of revolution?

This is very simple but I can't figure out what I'm doing wrong.

Let's say I'm rotating $f(x)=x^2$ around the y-axis. I have limits on the x-axis of a=1 and b=2. I want the volume under the curve, so I use the shells method, which gives me:

$2\pi \int_a^b f(x)x\delta x$ = $\frac{15}{2}\pi$

Now let's say I do it another way. I'll find the volume of the hollow cylinder with outer radius $b$ and inner radius $a$ and height f(b). This is $12\pi$. Now I want to subtract what's above the curve rotated around the y-axis.

I rotate function f around the y-axis, this time getting the volume above the curve. I can do this by integrating $f^{-1}(y)$ with limits $f(a)$ and $f(b)$. That gives me:

$\pi\int_{f(a)}^{f(b)}(f^{-1}(y))^2dy = \pi\int_1^4(\sqrt(y))^2dy = \pi\int_1^4y dy=\frac{15}{2}\pi$.

I'm sure at this point I've already made my mistake since the last answer gave me my final answer before I was expecting it.

Where's my mistake?

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There is a mild typo, you wrote $\pi\int_1^4x\, dy$ instead of $\pi\int_1^4y\, dy$, but that's not the problem, since you calculated as if it had been $y$.

In your second way, when you are trying to find out what to subtract, you are subtracting too much. You are using slicing to find the volume of the piece removed. So you need to find $$\int_1^4 \pi(y-1)\,dy.$$ This is because your slice also has a hole in it. The cross-section of the slice has a hole of radius $1$. So the area of the slice is $\pi((\sqrt{y})^2-1^2)$.

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Yeah thanks for pointing out the typo. In my head, I mix it up a little: I imagine 2 different $y$s, hence the confusion :) –  Korgan Rivera Mar 31 '12 at 22:51
Thanks, I see it now. I need to be careful about that. So I suppose it was just a coincidence that $2\pi \int_a^b f(x)x\delta x = \pi\int_{f(a)}^{f(b)}(f^{-1}(y))^2dy$. –  Korgan Rivera Mar 31 '12 at 22:55

In calculating the volume generated above the curve, you forgot that you have washers with outer radius $\sqrt{y}$ and inner radius $1$, not disks of radius $\sqrt{y}$: you should have

\begin{align*} \pi\int_1^4\left(\left(\sqrt{y}\right)^2-1^2\right)dy&=\pi\int_1^4(y-1)dy\\ &=\pi\left[\frac12y^2-y\right]_1^4\\ &=\pi\left(\frac{15}2-3\right)\\ &=\frac92\pi\;. \end{align*}

Now when you subtract this from $12\pi$ youl’ll get exactly what you expect.

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@David: Eep! Yes. –  Brian M. Scott Mar 31 '12 at 20:13