Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 60

Find the volume generated by rotating the region bounded by $y = e^x$, $x = 0$ and $y = \pi$ about $y$-axis. Use cylindrical shells method.
By using horizontal strips, notice that the strips have distance from the $x$-axis as $y$ such that if you rotate these lengths about the $x$-axis, you'll have a circumference of $c = 2 \pi y$. Also the height of the strips resembles the height of the cylinder as... $H = x_{\text{right}} - x_{\text{left}} = \ln y - 0 = \ln y$. Thus, the volume is...


$
\begin{equation}
\begin{aligned}
V &= \int^\pi_1 c(y) H(y) dy\\
\\
V &= \int^\pi_1 2\pi y(\ln y) dy\\
\\
V &= 2 \pi \int^\pi_1 y \ln y dy
\end{aligned}
\end{equation}
$


By using integration by parts, if we let $u = \ln y$ and $ dv = y dy$, then,
$\displaystyle du = \frac{1}{y} dy$ and $\displaystyle v = \frac{y^2}{2}$

$
\begin{equation}
\begin{aligned}
V &= 2\pi \int^\pi_1 y \ln y dy = uv - \int v du = 2 \pi \left[ \ln y \left( \frac{y^2}{2} \right) - \int \frac{y^2}{2} \left( \frac{dy}{y} \right) \right]\\
\\
&= 2 \pi \left[ \frac{y^2}{2} \ln y - \frac{1}{2} \frac{y^2}{2} \right]\\
\\
&= \frac{2 \pi y^2}{2} \left[ \ln y - \frac{1}{2} \right] \qquad = \pi y^2 \left[ \ln y - \frac{1}{2} \right]
\end{aligned}
\end{equation}
$


Evaluating for $y = 1$ to $y = \pi$

$
\begin{equation}
\begin{aligned}
&= \pi (\pi)^2 \left[ \ln \pi - \frac{1}{2} \right] - \pi (1)^2 \left[ \ln (1) - \frac{1}{2} \right] \\
\\
&= \pi^3 \left[ \ln \pi -\frac{1}{2} \right] + \frac{\pi}{2} \text{ cubic units}
\end{aligned}
\end{equation}
$