Single Variable Calculus, Chapter 6, 6.1, Section 6.1, Problem 40

Sketch the region defined by the inequalities $x - 2y^2 \geq 0, 1 - x - |y| \geq 0$ and find its area.







Based from the graph, it is more convenient to use horizontal strip to get the area bounded by the curves. So,

$\displaystyle A = \int^{y_2}_{y_1} (x_{\text{right}} - x_{\text{left}}) dy$

However, since we have inequality containing absolute value, the equation will quite change. So let's divide the area into two sub-region $A_1$ and $A_2$. Let $A_1$, we can get the point of intersection of the curve $x = 2y^2$ and $x = 1 - y$ by equating the two functions, so..


$
\begin{equation}
\begin{aligned}

2y^2 =& 1 - y
\\
\\
2y^2 + y - 1 =& 0

\end{aligned}
\end{equation}
$


By applying Quadratic Formula,

$\displaystyle y = \frac{1}{2}$ and $y = -1$

Thus, we have


$
\begin{equation}
\begin{aligned}

A_1 =& \int^{\frac{1}{2}}_0 [1 - y - (2y^2)] dy
\\
\\
A_1 =& \left[ y - \frac{y^2}{2} - \frac{2y^3}{3} \right]^{\frac{1}{2}}_0
\\
\\
A_1 =& \frac{7}{24} \text{ square units}

\end{aligned}
\end{equation}
$


Similarly from $A_2$, by using the functions $x = 2y^2$ and $x = 1 + y$


$
\begin{equation}
\begin{aligned}

2y^2 = 1 + y
\\
\\
2y^2 - y - 1 =& 0

\end{aligned}
\end{equation}
$


By applying Quadratic Formula,

$\displaystyle y = - \frac{1}{2}$ and $y = 1$

Thus, we have


$
\begin{equation}
\begin{aligned}

A_2 =& \int^0_{- \frac{1}{2}} [1 + y - 2y^2] dy
\\
\\
A_2 =& \left[ y + \frac{y^2}{2} - \frac{2y^3}{3} \right]^0_{- \frac{1}{2}}
\\
\\
A_2 =& \frac{7}{24} \text{ square units}

\end{aligned}
\end{equation}
$


Therefore, the total area is $\displaystyle A_1 + A_2 = \frac{7}{12}$ square units