Single Variable Calculus, Chapter 6, 6.3, Section 6.3, Problem 12

Use the shell method to find the volume generated by rotating the region bounded by the curves $x = 4y^2 - y^3, x = 0 $ 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 $H = 4y^2 - y^3$. So, we have..

$\displaystyle V = \int^b_a 2 \pi y f(y) dy$

To obtain the values of the upper and lower limit, we get the points of intersection of the curves..


$
\begin{equation}
\begin{aligned}

& 4y^2 - y^3 = 0
\\
& y^2 (4 - y) = 0
\\
& \text{We have,}
\\
& y = 0 \text{ and } y = 4

\end{aligned}
\end{equation}
$


Hence,


$
\begin{equation}
\begin{aligned}

V =& \int^4_0 2 \pi y(4y^2 - y^3 ) dy
\\
\\
V =& 2 \pi \int^4_0 (4y^3 - y^4) dy
\\
\\
V =& 2 \pi \left[ \displaystyle \frac{4y^4}{4} - \frac{y^5}{5} \right]^4_0
\\
\\
V =& \frac{512 \pi}{5} \text{ cubic units}

\end{aligned}
\end{equation}
$