find the area bounded by the curves y^2=2x+6 and x=y+1

Answer 1

The curves intersect at (-1,-2) and (5,4)

Horizontal strips of width dy are the best here, so

A = ∫[-2,4] (y+1)-(y^2-6)/2 dy = 18

Using vertical strips is a bit trickier, because the boundary changes at x = -1

A = ∫[-3,-1] √(2x+6) - (-√(2x+6)) dx
+ ∫[-1,5] √(2x+6)-(x-1) dx
= 38/3 + 16/3 = 54/3 = 18

Answer 2

Thank you!