Integral Homework Question I've tried this problem twice now and I only have one more try before I get it wrong; I was wondering if anyone could help me with it.
"Find the volume of the solid generated by revolving the region enclosed by the graphs of y=e^(x/2), y=1, and x=ln(3) about the x-axis."
We are supposed to use the shell method, and I believe I set the problem up correctly. I would appreciate it if anyone could walk me through the problems or just start it for me to see if I am doing the problem wrong.
I set up the problem as [Integrate]1 -> sqrt(3): 2pi(y)(ln(3)-2ln(y))dy.
The answer I got was ((5ln(3)-2)pi)/2.
EDIT: Answered.
 A: Let's slice it first.  The curve $e^{x/2}$ and $y=1$ intersect at $x=0$, and the right boundary is at $x=\ln 3$.
So we'll add up the volumes of all of the washers with outer radius $e^{x/2}$ and inner radius $1$:
$$V = \pi \int_0^{\ln 3} (e^{x} - 1) dx = \left. \pi (e^{x} - x)\right|_0^{\ln 3} = \pi (3 - \ln 3 - 1) = \pi (2 - \ln 3).$$
Now for the shell method.  The inverse of the curve is $x = 2 \ln y$, but we're looking for the volume to the right of this curve, and to the left of $x = \ln 3.$  At $x = \ln 3$, $y = \sqrt{3}$, and at $x=0, y=1$, so that's our outer limit of integration.
Then, the volume is the sum of all of the shells:
$$V = 2 \pi \int_1^{\sqrt{3}} y (\ln 3 - 2 \ln y) dy = 2\pi \ln 3 \int_1^{\sqrt{3}} y \text{ d}y  - 4 \pi  \int_1^{\sqrt{3}}  y \ln y \text{ d}y= \pi (2 - \ln 3).$$
A: Your setup looks just fine, so I suppose some mistake was probably made in the evaluation. This isn't surprising, though, as the integral using this approach isn't very straightforward to evaluate. I'll go ahead and work it through so you can see how.
First off, we can separate it into two integrals and pull out the constants to make our job easier: $$\begin{align}\int_1^{\sqrt3}2\pi y(\ln 3-2\ln y)\,dy &= \int_1^{\sqrt3}\left(2\pi y\ln 3-4\pi y\ln y\right)\,dy\\ &= 2\pi\ln 3\int_1^{\sqrt 3}y\,dy-4\pi\int_1^{\sqrt3}y\ln y\,dy\\ &= 2\pi\ln 3\left[\frac12y^2\right]_{y=1}^{\sqrt 3}-4\pi\int_1^{\sqrt3}y\ln y\,dy\\ &= 2\pi\ln 3\left[\frac32-\frac12\right]-4\pi\int_1^{\sqrt3}y\ln y\,dy\\ &= 2\pi\ln 3-4\pi\int_1^{\sqrt3}y\ln y\,dy\end{align}$$
At this point, you can use integration by parts to find an antiderivative of $y\ln y$:
$$\begin{align}\int y\ln y\,dy &= \int\ln y\cdot d\left(\frac12y^2\right)\\ &= \ln y \cdot \frac12y^2-\int\left(\frac12y^2\right)\cdot d(\ln y)\\ &= \frac12y^2\ln y-\frac12\int y^2\cdot \frac1y\,dy\\ &= \frac12y^2\ln y-\frac12\int y\,dy\\ &= \frac12 y^2\ln y-\frac14y^2\end{align}$$ Hence, $$\begin{align}\int_1^{\sqrt3}y\ln y\,dy &= \left[\frac12 y^2\ln y-\frac14 y^2\right]_{y=1}^{\sqrt3}\\ &= \left[\left(\frac32\ln\sqrt3-\frac34\right)-\left(-\frac14\right)\right]\\ &= \frac32\ln 3^{1/2}-\frac12\\ &=\frac34\ln3-\frac12,\end{align}$$ and so $$\begin{align}\int_1^{\sqrt3}2\pi y(\ln 3-2\ln y)\,dy &= 2\pi\ln 3-4\pi\int_1^{\sqrt3}y\ln y\,dy\\ &= 2\pi\ln 3-3\pi\ln 3+2\pi\\ &=2\pi-\pi\ln 3\\ &= \pi(2-\ln 3).\end{align}$$
