Finding the Area Between 2 Curves -- One Intersection Point?

Question

I'm struggling to find the solution to the following problem: Find the area of the region enclosed by the graphs of $y^2=x+8$ and $y^2=4-x$.

My initial thought was to square root both equations and set them equal to each other. Then square the $\sqrt{x+8}=\sqrt{4-x}$. However, I am only getting one intersection point, $x=-2$.

How do I proceed with the problem if I only have one boundary?

Answer 1

There are in fact two points of intersection. Remember that a square root is both positive and negative. The points are: (spoiler)

$(-2, \sqrt{6})$ and $(-2, -\sqrt{6})$

Graph for reference:

Note that you must always first draw a rough sketch of the functions whose area you're supposed to compute. Drawing the rough sketch would immediately tell you that both functions are symmetric about the x-axis (which was also evident from their original equations). So, there must be even number of points of intersection.

Answer 2

When $x=-2$ you have $y^2=6,$ so $y=\pm \sqrt{6}.$