The curves intersect at (-1,2) and (-1,-2)
So, using horizontal strips of width dy, the area is
a = ∫[-2,2] (3-y^2)-(y^2-5) dy
Using vertical strips, we have to split the region in two at the intersections, and then symmetry helps:
a = ∫[-5,-1] 2√(x+5)dx
+ ∫[-1,3] 2√(3-x)dx
Consider the region enclosed by the graphs of x=y^2-5 and x=3-y^2
a)Express the area of this region by setting up an integral with respect to x
b) Express the area of this region by setting up an integral with respect to y
c) Find the area of this region by evaluating one of the definite integrals found above
1 answer