Single Variable Calculus, Chapter 6, 6.3, Section 6.3, Problem 16

Use the shell method to find the volume generated by rotating the region bounded by the curves $y = \sqrt{x}, y = 0, x = 1$ about the $x = -1$. Sketch the region and a typical shell.

If we use a horizontal strips, notice that the distance of the strips from the line $x = -1$ is $x + 1$. If we revolve this length about $x = -1$, you'll get the circumference $C(x) = 2 \pi (x + 1)$. Also, notice that the height of the strips resembles the height of the cylinder as $H = y_{\text{upper}} - y_{\text{lower}} = \sqrt{x} - 0$. Thus, we have..

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


$
\begin{equation}
\begin{aligned}

V =& \int^1_0 2 \pi (x + 1) (\sqrt{x}) dx
\\
\\
V =& 2 \pi \int^1_0 (x^{\frac{3}{2}} + x^{\frac{1}{2}}) dx
\\
\\
V =& 2 \pi \left[ \displaystyle \frac{x^{\frac{5}{2}}}{\displaystyle \frac{5}{2}} + \frac{x^{\frac{3}{2}}}{\displaystyle \frac{3}{2}} \right]^1_0
\\
\\
V =& \frac{32 \pi}{15} \text{ cubic units}

\end{aligned}
\end{equation}
$