Single Variable Calculus, Chapter 6, 6.3, Section 6.3, Problem 14

Use the shell method to find the volume generated by rotating the region bounded by the curves $x + y = 3, x = 4 - (y - 1)^2$ 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 $H = x_{\text{right}} - x_{\text{left}} = 4 - (y - 1)^2 - (3 - y)$. Thus, we have..

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

In order to obtain the valued of the upper and lower limits, we get the points of intersection of the curves.


$
\begin{equation}
\begin{aligned}

& 3 - y = 4 - (y - 1)^2
\\
& 3 - y = 4 - y^2 + 2y - 1
\\
& y^2 - 3y = 0
\\
& y(y - 3) = 0
\\
& \text{We have,}
\\
& y = 0 \text{ and } y = 3

\end{aligned}
\end{equation}
$


Thus, we have


$
\begin{equation}
\begin{aligned}

V =& \int^3_0 2 \pi [4 - (y - 1)^2 - (3 - y)] dy
\\
\\
V =& 2 \pi \int^3_0 (4 - y^2 + 2y - 1 - 3 + y) dy
\\
\\
V =& 2 \pi \int^3_0 (-y^3 + 3y^2) dy
\\
\\
V =& 2 \pi \left[ \frac{-y^4}{4} + \frac{3y^3}{3} \right]^3_0
\\
\\
V =& \frac{27 \pi}{2} \text{ cubic units}

\end{aligned}
\end{equation}
$