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.
Best Answer
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.
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.