Single Variable Calculus, Chapter 6, 6.3, Section 6.3, Problem 8

Use the shell method to find the volume generated by rotating the region bounded by the curves $y = \sqrt{x}, y = x^2$ about the $y$-axis. Sketch the region and a typical shell








If we use a vertical strips, notice that the distance of these strips from the $y$-axis is $x$. If we revolve this length about the $y$-axis, you'll get the circumference $C = 2 \pi x$. Also, notice that the height of the strips resembles the height of the cylinder as $H y_{\text{upper}} - y_{\text
{lower}} = \sqrt{x} - x^2$. So, we have..

$\displaystyle V = \int^b_a C(x) H(x) dx$

The values of the upper and lower limits can be obtained by getting the points of intersection.


$
\begin{equation}
\begin{aligned}

& \sqrt{x} = x^2
\\
\\
& x^{(2 - \frac{1}{2})} = 1
\\
\\
& x^{\frac{3}{2}} = 1
\\
\\
& x = 1

\end{aligned}
\end{equation}
$


Hence,


$
\begin{equation}
\begin{aligned}

V =& \int^1_0 2 \pi x (\sqrt{x} - x^2) dx
\\
\\
V =& 2 \pi \int^1_0 (x^{\frac{3}{2}} - x^3) dx
\\
\\
V =& 2 \pi \left[ \displaystyle \frac{x^{\frac{5}{2}}}{\displaystyle \frac{5}{2}} - \frac{x^4}{4} \right]^1_0
\\
\\
V =& \frac{3 \pi}{10} \text{ cubic units}

\end{aligned}
\end{equation}
$