# Calculus - Integrals

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

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1. 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

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posted by Steve

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