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

  1. 0
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asked by Beth
  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

    posted by Steve

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