Single Variable Calculus, Chapter 6, 6.3, Section 6.3, Problem 18

Use the shell method to find the volume generated by rotating the region bounded by the curves $y = x^2, y = 2 - x^2$ about the $x = 1$. Sketch the region and a typical shell.








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

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

In order to get the values of the upper and lower limits, we simply determine the points of intersection of the curves...


$
\begin{equation}
\begin{aligned}

x^2 &= 2 - x^2
\\
2x^2 &= 2
\\
x^2 &= 1
\\
x &= \pm 1

\end{aligned}
\end{equation}
$


Therefore, we have..


$
\begin{equation}
\begin{aligned}

V =& \int^1_{-1} 2 \pi (1 - x) (2 - x^2 - (x^2)) dx
\\
\\
V =& 2 \pi \int^1_{-1} (2 - 2x^2 - 2x + 2x^3) dx
\\
\\
V =& 2 \pi \left[ 2x - \frac{2x^3}{3} - \frac{2x^2}{2} + \frac{2x^4}{4} \right]^1_{-1}
\\
\\
V =& \frac{16 \pi}{3} \text{ cubic units}

\end{aligned}
\end{equation}
$