Volume from revolving $e^x$ around $y$-axis

Question

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.

Thanks.

Answer 1

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

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

The arrows indicate the radii of some typical disks.

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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 $y=e^x\iff x=\ln y$:

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

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