notice that one parabola opens up, the other down.
their intersection .....
x^2 - c^2 = c^2 - x^2
2x^2 = 2c^2
x^2 = c^2
x = ± c ---- y = 0 in both
they intersect at (-c,0) and (c,0)
using symmetry
area = 2∫(2c^2 - 2x^2) dx from 0 to c
= 2(2c^2 x - (2/3)x^3) | from 0 to c
= 2( 2c^3 - (2/3)c^2 - 0)
= 2(4/3 c^3)
so (8/3)c^3 = 13
8c^3 = 39
c = ∛39 /2 = appr 1.7
check my algebra and arithmetic, was expecting a "nicer" answer.
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.
1 answer