So, the area between the two functions is 72 units squared.

[tmt1]

I need to find the are between $$y1 = 12 - x^2$$ and $$ y2 = x^2 - 6$$.

Since y1 is greater, I subtract y2 from y1 getting:

$$ \int 18 - 2x^2$$ which is $$18x - 2x^3 / 3$$,

The intersecting points are $$x = -3 and x= 3$$.

So I find $$18x - 2x^3 / 3 from x = 3 to x = -3$$(I'm trying to figure out how to do this in latex)

which just equals 108, however the answer is 72, so what am I doing wrong here?

 

[MarkFL]

Use the underscore character for subscripting:

$$y_1=12-x^2$$

$$y_2=x^2-6$$

I prefer to find the limits of integration first:

$$12-x^2=x^2-6$$

$$2x^2=18\implies x=\pm3$$

We have two even functions, and limits that are symmetric about the $y$-axis, so we can apply the even function rule, to give the enclosed area $A$ as:

$$A=2\int_0^3 18-2x^2\,dx=4\int_0^3 9-x^2\,dx=4\left[9x-\frac{x^3}{3}]\right]_0^3=4\left(27-9\right)=72$$