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

Use the shell method to find the volume generated by rotating the region bounded by the curves $y = 3 + 2x - x^2, x + y = 3$ 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 $H = 3 + 2x - x^2 - ( 3 - x)$. Thus, we have..

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

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


$
\begin{equation}
\begin{aligned}

& 3 + 2x - x^2 = 3 - x
\\
\\
& 3x - x^2 = 0
\\
\\
& x(3 - x) = 0
\\
\\
& \text{ we have}
\\
\\
& x = 0 \text{ and } x = 3
\end{aligned}
\end{equation}
$


Hence,


$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$