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SOLUTION 29: Compute the area of the region enclosed by the graphs of the equations y=x2−2x+2 (or x=1±√y−1)
and y=2x+2 (or x=12y−1).Beginbyfindingthepointsofintersectionofthethetwographs.From y= x^2-2x+2 and y=2x+2 wegetthatx2−2x+2=2x+2 ⟶x2−4x=0 ⟶x(x−4)=0 ⟶ x=0 (and y=2) or x=4 (and y=10)$
Now see the given graph of the enclosed region.
a.) Using vertical cross-sections to describe this region, we get that
0≤x≤4 and x2−2x+2≤y≤2x+2 ,
so that the area of this region is
AREA=∫40(Top − Bottom) dx
=∫40((2x2−2x+2)−(2x+2)) dx
=∫40(2x2−4x) dx
=(23x3−2x2)|40
=(23(4)3−2(4)2)−(23(0)3−2(0)2)
=(1283−32)−(0)
=1283−963
=323
b.) Using horizontal cross-sections to describe this region, which is made up of two smaller regions, we get that
1≤y≤2 and 1−√y−1≤x≤1+√y−1
in addition to
2≤y≤10 and 12y−1≤x≤1+√y−1 ,
so that the area of this region is
AREA=∫21(Right − Left) dy+∫102(Right − Left) dy
=∫21((1+√y−1)−(1−√y−1)) dy+∫102((1+√y−1)−(12y−1)) dy
=∫212⋅√y−1 dy+∫102(2+√y−1−12y) dy
=(43(y−1)3/2)|21+(2y+23(y−1)3/2−14y2)|102
=(43((2)−1)3/2−43((1)−1)3/2)+(2(10)+23((10)−1)3/2−14(10)2)−(2(2)+23((2)−1)3/2−14(2)2)
=(43(1)3/2−43(0)3/2)+(20+23(9)3/2−25)−(4+23(1)3/2−1)
=(43)+(23(27)−5)−(3+23)
=43+13−3−23
=23+303
=323
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