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 90.

Answer 1

Did you make a sketch?
Did you notice that one parabola is the reflection of the other in the x-axis.
Lots of symmetry here
The 2 axes cut the region into 4 equal parts , each one would then be 22.5
Did you notice the x-intercept on the right side is (c,0) ?

okay then,
∫(c^2 - x^2) dx from 0 to c = 22.5

the integral is c^2 x - (1/3)x^3
so
c^3 - (1/3)c^3 = 22.5
(2/3) c^3 = 45/2
c^3 = 135/4 or 270/8 , (looking for perfect cubes, only got one)
c = (270/8)^(1/3)
= 3(cuberoot(10)/2 , (270 = 270x10 , 27 is a perfect cube)

I get appr 3.23