SOLUTION 29: Compute the area of the region enclosed by the graphs of the equations $ y= x^2-2x+2\ \ (or \ \ x=1 \pm \sqrt{y-1})$
and $ y= 2x+2 \ \ (or \ \ \displaystyle{ x= \frac{1}{2}y-1 } ). Begin by finding the points of intersection of the the two graphs.
From $ y= x^2-2x+2 $ and $ y=2x+2 $ we get that
$$ x^2-2x+2 = 2x+2 \ \ \longrightarrow $$
$$ x^2-4x = 0 \ \ \longrightarrow $$
$$ x(x-4) = 0 \ \ \longrightarrow \ \ x=0 \ \ ( and \ \ y=2) \ \ or \ \ x=4 \ \ (and \ \ y=10) $$
Now see the given graph of the enclosed region.
a.) Using vertical cross-sections to describe this region, we get that
$$ 0 \le x \le 4 \ \ and \ \ x^2-2x+2 \le y \le 2x+2 \ , $$
so that the area of this region is
$$ AREA = \displaystyle{ \int_{0}^{4} \Big( Top \ - \ Bottom \Big) \ dx } $$
$$ = \displaystyle{ \int_{0}^{4} \Big( (2x^2-2x+2) - (2x+2) \Big) \ dx } $$
$$ = \displaystyle{ \int_{0}^{4} \Big( 2x^2-4x \Big) \ dx } $$
$$ \displaystyle { = \Big( \frac{2}{3}x^3 - 2x^2 \Big) \Big\vert_{0}^{4} } $$
$$ \displaystyle { = \Big( \frac{2}{3}(4)^3 - 2(4)^2 \Big) - \Big( \frac{2}{3}(0)^3 - 2(0)^2 \Big) } $$
$$ \displaystyle { = \Big( \frac{128}{3} - 32 \Big) - \Big( 0 \Big) } $$
$$ \displaystyle { = \frac{128}{3} - \frac{96}{3} } $$
$$ \displaystyle { = \frac{32}{3} } $$
b.) Using horizontal cross-sections to describe this region, which is made up of two smaller regions, we get that
$$ 1 \le y \le 2 \ \ and \ \ 1 - \sqrt{y-1} \le x \le 1 + \sqrt{y-1} $$
in addition to
$$ 2 \le y \le 10 \ \ and \ \ \displaystyle{ \frac{1}{2}y-1 } \le x \le 1 + \sqrt{y-1} \ , $$
so that the area of this region is
$$ AREA = \displaystyle{ \int_{1}^{2} ( Right \ - \ Left ) \ dy + \int_{2}^{10} ( Right \ - \ Left ) \ dy } $$
$$ = \displaystyle{ \int_{1}^{2} \Big( \Big( 1 + \sqrt{y-1} \Big) - \Big( 1 - \sqrt{y-1} \Big) \Big) \ dy
+ \int_{2}^{10} \Big( \Big( 1 + \sqrt{y-1} \Big) - \Big( \frac{1}{2}y-1\Big) \Big) \ dy } $$
$$ = \displaystyle{ \int_{1}^{2} 2 \cdot \sqrt{y-1} \ dy
+ \int_{2}^{10} \Big( 2 + \sqrt{y-1} - \frac{1}{2}y \Big) \ dy } $$
$$ \displaystyle{ = \Big( \frac{4}{3}(y-1)^{3/2} \Big) \Big\vert_{1}^{2}
+ \Big( 2y + \frac{2}{3}(y-1)^{3/2} - \frac{1}{4}y^2 \Big) \Big\vert_{2}^{10} } $$
$$ \displaystyle{ = \Big( \frac{4}{3}((2)-1)^{3/2} - \frac{4}{3}((1)-1)^{3/2} \Big)
+ \Big( 2(10) + \frac{2}{3}((10)-1)^{3/2} - \frac{1}{4}(10)^2 \Big)
- \Big( 2(2) + \frac{2}{3}((2)-1)^{3/2} - \frac{1}{4}(2)^2 \Big) } $$
$$ \displaystyle{ = \Big( \frac{4}{3}(1)^{3/2} - \frac{4}{3}(0)^{3/2} \Big)
+ \Big( 20 + \frac{2}{3}(9)^{3/2} - 25 \Big)
- \Big( 4 + \frac{2}{3}(1)^{3/2} - 1 \Big) } $$
$$ \displaystyle{ = \Big( \frac{4}{3} \Big)
+ \Big( \frac{2}{3}(27) - 5 \Big)
- \Big( 3 + \frac{2}{3} \Big) } $$
$$ \displaystyle{ = \frac{4}{3} + 13 - 3 - \frac{2}{3} } $$
$$ \displaystyle{ = \frac{2}{3} + \frac{30}{3} } $$
$$ \displaystyle{ = \frac{32}{3} } $$
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