Find c > 0 such that the area of the region enclosed by the parabolas y = x^2 - c^2 and y = c^2 -x^2 is 13. Crazy integral.... HeeeLLLPP!?

First, we let
f(x)=x^2-c^2
g(x)=c^2-x^2=-f(x)
Area=integral of (g(x)-f(x))=2(c^2-x^2)

We need to find the limits of integration, which are the intersection points of f(x) and g(x):
f(x)=g(x) => x^2-c^2=c^2=x^2 => x=+/- C
Therefore area A=integral 2(c^2-x^2) limits are from -c to +c.
The required area is the tip parts of the parabolas, back to back.

We still don't know what c should be, so we make use of the remaining information:
A=13.
First, we will find indefinite integral A in terms of x and c:
A(x,c)=2x(c^2-x^2/3)
and we know that
total area = A(c,c)-A(-c,c) = 13
Substituting x=c and x=-c,
2c(2c^2/3)-2(-c)(2(-c)^2/3)=13
and simplifying
8c^3/3=13
=>
c=sqrt(39)/2

Check: set c=sqrt(39)/2
Area A = integrate(2c^2x-2x^3/3)dx from x=-c to x=c
gives A=13, so the answer is correct.

Hints:
* What value(s) of x yield y = 0?
* What is the value of y when x = c? When x = -c?
* Make a sketch.
* What is the integral of a constant with respect to x?