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In school, I had a problem something like this:

A region R is bounded by $x$-axis, $y$-axis, $x = 3$, and $y = e^x$. What is the volume of the solid produced by revolving it around the $y$-axis.

To solve this, using the washer method, I thought I would have to do $$\pi\int_1^{e^3}[3^2-\ln(y)^2]\,dy+\pi{(3)^2}\cdot1$$ to get the area above where $e^x$ intersects the axis and then the area of the cylinder from $0$ to $1$.

However, my teacher said the answer would be only $$\pi\int_0^{e^3}[3^2-\ln(y)^2]\,dy.$$

I felt that this integral would also be getting the area of the curve that was not bounded by the $y$-axis from $0$ to $1$, but my teacher didn't really have a good explanation for the question.

Which answer is correct? If it is the teacher's answer, please explain why my thought process was wrong.


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Your answer would be correct if you revolved $R$ about the $y$-axis (revolving $R$ about the $x$-axis leads to a disc method problem where you integrate with respect to $x$) and if you multiply your integral by $\pi$. If it is the case that you're revolving about the $y$-axis, your teacher's answer is not correct. – David Mitra Jan 12 '13 at 15:00
@DavidMitra Sorry, I typed this on my phone. I intended to have a pi, and it is around the y-axis. I have changed the question to reflect that. If you could post your comment as an answer, I could accept it. – Vishnu Jan 12 '13 at 15:18
I'll just refer you to JohnD's answer. I would have just told you to draw the picture, and note that the function giving the "inner radius" of the washer is different depending on whether $y\in[0,1]$ (in which case it's $0$), or $y\in[1,e^3]$ (in which case it's $\ln y$). Of course, it's easier to just note, as you did, that you have a cylinder at the bottom and a solid of revolution on top whose volume can be expressed nicely as an integral. – David Mitra Jan 12 '13 at 15:34
up vote 3 down vote accepted

If you revolve about the x axis, you get disks not washers, since the "inner radius" is $0$. Thus,

$$ \text{volume}=\pi\int_a^b (R(x))^2\,dx=\pi\int_0^3 (e^{x})^2\,dx={\pi(e^6-1)\over 2} $$

Mathematica graphics

The arrows indicate the radii of some typical disks.

However, if you revolve about the $y$ axis instead, the you do get washers and you should break the region up into two parts based on when the inner radius changes from $0$ to following the curve $y=e^x\iff x=\ln y$:

$$ \text{volume}=\pi\int_0^1 (3^2-0^2)\,dy+\pi\int_1^{e^3} [(3^2-(\ln(y))^2]\,dy =9 \pi +\pi\left(4 e^3-7\right). $$

Mathematica graphics

The arrows indicate the inner and outer radii of some typical washers.

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Where is the $1^2$ coming from when revolving around the y-axis? Why is it not just $\pi\int_0^1{3^2}dy$? – Vishnu Jan 12 '13 at 15:30
You all caught me while editing. Fixed now. ;-) – JohnD Jan 12 '13 at 15:32
OK, thanks for the thorough answer – Vishnu Jan 12 '13 at 15:34

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