Single Variable Calculus, Chapter 6, 6.3, Section 6.3, Problem 10

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








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

$\displaystyle V = \int^b_a C(y) H(y) dy$


$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$