Single Variable Calculus, Chapter 6, 6.1, Section 6.1, Problem 18

Sketch the region enclosed by the curves $y = 8 - x^2$, $y = x^2$, $x = -3$ and $x = 3$. Then find the area of the region.


By using vertical strips
$\displaystyle A = \int^{x_2}_{x_1} \left(y_{\text{upper}} - y_{\text{lower}} \right) dx$
In order to get the values of the upper and lower limits, we equate the two functions to get its point of intersection. Thus


$
\begin{equation}
\begin{aligned}
8 - x^2 &= x^2\\
\\
2x^2 &= 8 \\
\\
x^2 &= 4
\end{aligned}
\end{equation}
$

we have, $x = 2$ and $ x = -2$
Let's divide the shaded region on three parts. Let $A_1$, $A_2$ and $A_3$ be the area of left most part, middle part and the right most part respectively. Thus,

$
\begin{equation}
\begin{aligned}
A_1 &= \int^{-2}_{-3} \left[ x^2 - \left( 8 - x^2 \right)\right] dx\\
\\
A_1 &= \int^{-2}_{-3} \left[ 2x^2 - 8 \right] dx\\
\\
A_1 &= \left[ \frac{2x^3}{3} - 8x \right]^{-2}_{-3}\\
\\
A_1 &= \frac{14}{3} \text{ square units}

\end{aligned}
\end{equation}
$

For the middle part,

$
\begin{equation}
\begin{aligned}
A_2 &= \int^{2}_{-2} \left[ x^2 - \left( 8 - x^2 \right)\right] dx\\
\\
A_2 &= \int^2_{-2} \left[ 8 - 2x^2 \right]dx \\
\\
A_2 &= \left[ 8x - \frac{2x^3}{3} \right]^2_{-2}\\
\\
A_2 &= \frac{64}{3} \text{ square units}

\end{aligned}
\end{equation}
$


For the right most part,

$
\begin{equation}
\begin{aligned}
A_3 &= \int^3_2 \left[ x^2 - \left( 8 - x^2 \right)\right] dx\\
\\
A_3 &= \int^3_2 \left[ 2x^2 - 8 \right]\\
\\
A_3 &= \left[ \frac{2x^3}{3} -8x \right]^3_2\\
\\
A_3 &= \frac{14}{3} \text{ square units}
\end{aligned}
\end{equation}
$



Therefore, the area of the entire region is $\displaystyle A_1 + A_2 + A_3 = \frac{14}{3} + \frac{64}{3} + \frac{14}{3} = \frac{92}{3}$ square units