Answer
The area of the region is $243/8$.
Work Step by Step
$y^2-4x=4$ and $4x-y=16$
- For $y^2-4x=4$, we can rewrite $$4x=y^2-4$$ $$x=\frac{y^2}{4}-1$$
- For $4x-y=16$, we can rewrite $$4x=y+16$$ $$x=\frac{y}{4}+4$$
1) Draft the graph:
The graph is drafted below. We would use integration with respect to $y$ here.
2) Find the limits of integration:
To find the limits of integration, we find intersection points with respect to $y$.
$$\frac{y^2}{4}-1=\frac{y}{4}+4$$ $$y^2-4=y+16$$ $$y^2-y-20=0$$ $$y=-4\hspace{1cm}\text{or}\hspace{1cm}y=5$$
So the upper limit is $5$ and the lower limit is $-4$.
3) Find the area:
The region from $y=-4$ to $y=5$ is bounded on the right by $x=\frac{y}{4}+4$ and on the left by $x=\frac{y^2}{4}-1$. Therefore, the area of the region is $$A=\int^5_{-4}[(\frac{y}{4}+4)-(\frac{y^2}{4}-1)]dy=\int^5_{-4}\Big(-\frac{y^2}{4}+\frac{y}{4}+5\Big)dy$$ $$A=-\frac{y^3}{12}+\frac{y^2}{8}+5y\Big]^5_{-4}$$ $$A=\Big(-\frac{125}{12}+\frac{25}{8}+25\Big)-\Big(\frac{16}{3}+2-20\Big)$$ $$A=\frac{425}{24}-\Big(-\frac{38}{3}\Big)=\frac{243}{8}$$
