Single Variable Calculus, Chapter 6, 6.1, Section 6.1, Problem 20

Sketch the region enclosed by the curves $4x + y^2 = 12$, $y = x$. Then find the area of the region.


By using vertical strips
$\displaystyle A = \int^{y_2}_{y_1} \left(x_{\text{right}} - x_{\text{left}} \right) dx$
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}
\frac{12-y^2}{4} &= y\\
\\
12 - y^2 &= 4y\\
\\
y^2 + 4y - 12 &= 0
\end{aligned}
\end{equation}
$


By applying Quadratic Formula,
$y = 2$ and $y = -6$
Therefore,

$
\begin{equation}
\begin{aligned}
A &= \int^2_{-6} \left[ \left( \frac{12-y^2}{4}\right) - y \right] dy\\
\\
A &= \int^2_{-6} \left[ \frac{12-y^2 - 4y}{4} \right] dy\\
\\
A &= \left[ 3y - \frac{y^3}{3(4)} - \frac{y^2}{2} \right]^2_{-6}\\
\\
A &= \frac{64}{3} \text{ square units}
\end{aligned}
\end{equation}
$