Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 62

Find the volume obtained by rotating the region bounded by $y = \sin^2 x$ and $y = 0$ from $0 \leq x \leq \pi$ about $x$-axis.




By using vertical strips, if you slice the figure, you'll get a cross sections of circle with radius $r = y_{\text{upper}} - y_{\text{lower}} = \sin^x - 0 = \sin^2 x$. So, the cross sectional area is computed by $A = \pi r^2 = \pi \left( \sin^2 x \right)^2$. Thus, the volume is...

$
\begin{equation}
\begin{aligned}
V = \int^b_a A (x) dx &= \int^\pi_0 \pi \left( \sin^2 x \right)^2 && \text{ recall that } \sin^2 x = \frac{1 - \cos (2x)}{2}\\
\\
&= \pi \int^\pi_0 \left[ \frac{1 - \cos (2x) }{2} \right]^2 dx\\
\\
&= \frac{\pi}{4} \int^\pi_0 \left[ 1 - 2 \cos (2x) + \cos^2 (2x) \right] dx && \text{ recall that } \cos^2(x) = \frac{1 + \cos (2x) }{2}\\
\\
&= \frac{\pi}{4} \int^\pi_0 \left[ 1 - 2 \cos (2x) + \left[ \frac{1+\cos(2(2x))}{2} \right]\right]dx \\
\\
&= \frac{\pi}{4} \int^\pi_0 \left[ \frac{3}{2} - 2 \cos (2x) + \frac{\cos(4x)}{2} \right] dx\\
\\
&= \frac{\pi}{4} \left[ \int^\pi_0 \frac{3}{2}dx - 2 \int^\pi_0 \cos (2x) dx + \frac{1}{2} \int^\pi_0 \cos(4x) dx \right]\\
\\
&= \frac{\pi}{4} \left[ \left( \frac{3}{2} x \right)^\pi_0 - 2 \left( \frac{1}{2} \right) \left[ \sin (2x) \right]^\pi_0 + \frac{1}{2} \left( \frac{1}{4} \right) \left[ \sin (4x) \right]^\pi_0 \right]\\
\\
&= \frac{3 \pi^2}{8} \text{ cubic units}
\end{aligned}
\end{equation}
$