Find the area enclosed by the equations x=16-y^2 and x=y^2-16

Answer 1

I don't think the equations enclose any area. But their graphs do. Gotta be precise when talking and doing math!

The curves intersect at y=±4, so the area between the curves is just

∫[-4,4] (16-y^2)-(y^2-16) dy

which, taking advantage of the symmetry is

4∫[0,4] 16-y^2 dy
= 4(16y - 1/3 y^3) [0,4]
= 4(64 - 64/3)
= 512/3

You could also take vertical strips instead of horizontal, so the area is

4∫[0,16] √(16-x) dx
= 8/3 (16-x)^(3/2)
= 512/3