SOLUTION 25: Compute the area of the region enclosed by the graphs of the equations $ x=y^2 \ \ (or \ \ y=\sqrt{x} \ \ and \ \ y=-\sqrt{x})$
and $ x=y+2 \ \ (or \ \ y=x-2) $ . Begin by finding the points of intersection of the the two graphs. From $ x=y^2 $ and
$ x = y+2 $ we get that
$$ y^2 = y+2 \ \ \longrightarrow $$
$$ y^2-y-2 = 0 \ \ \longrightarrow $$
$$ (y-2)(y+1) = 0 \ \ \longrightarrow \ \ y=2 \ \ (and \ x=4) \ \ or \ \ y=-1 \ \ (and \ x=1) $$
Now see the given graph of the enclosed region.
a.) Using vertical cross-sections to describe this region, we get that
$$ 0 \le x \le 1 \ \ and \ \ -\sqrt{x} \le y \le \sqrt{x} \ $$
in addition to
$$ 1 \le x \le 4 \ \ and \ \ x-2 \le y \le \sqrt{x} \ , $$
so that the area of this region is
$$ AREA = \displaystyle{ \int_{0}^{1} ( Top \ - \ Bottom ) \ dx + \int_{1}^{4} ( Top \ - \ Bottom \ dx } $$
$$ = \displaystyle{ \int_{0}^{1} (\sqrt{x} - (- \sqrt{x}) ) \ dx + \int_{1}^{4} (\sqrt{x} - (x-2)) \ dx } $$
$$ = \displaystyle{ \int_{0}^{1} 2 \sqrt{x} \ dx + \int_{1}^{4} (\sqrt{x} - x + 2) \ dx } $$
$$ \displaystyle { = \Big( \frac{4}{3} x^{3/2} \Big) \Big\vert_{0}^{1}
+ \Big( \frac{2}{3} x^{3/2} - \frac{x^2}{2} + 2x \Big) \Big\vert_{1}^{4} } $$
$$ \displaystyle { = \Big( \frac{4}{3} (1)^{3/2} - \frac{4}{3} (0)^{3/2} \Big)
+ \Big( \frac{2}{3} (4)^{3/2} - \frac{(4)^2}{2} + 2(4) \Big)
- \Big( \frac{2}{3} (1)^{3/2} - \frac{(1)^2}{2} + 2(1) \Big) \Big) } $$
$$ \displaystyle { = \Big( \frac{4}{3} \Big) + \Big( \frac{16}{3} \Big) - \Big( \frac{2}{3} + \frac{3}{2} \Big) } $$
$$ \displaystyle { = \frac{18}{3} - \frac{3}{2} } $$
$$ \displaystyle { = 6 - \frac{3}{2} } $$
$$ \displaystyle { = \frac{12}{2} - \frac{3}{2} } $$
$$ \displaystyle { = \frac{9}{2} } $$
b.) Using horizontal cross-sections to describe this region, we get that
$$ -1 \le y \le 2 \ \ and \ \ y^2 \le x \le y+2 \ , $$
so that the area of this region is
$$ AREA = \displaystyle{ \int_{-1}^{2} ( Right \ - \ Left ) \ dy } $$
$$ = \displaystyle{ \int_{-1}^{2} ( (y+2) - y^2) \ dy } $$
$$ \displaystyle { = \Big( \frac{y^2}{2} + 2y - \frac{y^3}{3} \Big) \Big\vert_{-1}^{2} } $$
$$ \displaystyle { = \Big( \frac{(2)^2}{2} + 2(2) - \frac{(2)^3}{3} \Big)
- \Big( \frac{(-1)^2}{2} + 2(-1) - \frac{(-1)^3}{3} \Big) } $$
$$ \displaystyle { = \Big( 6 - \frac{8}{3} \Big)
- \Big( \frac{1}{2} - 2 + \frac{1}{3} \Big) } $$
$$ \displaystyle { = 6 - \frac{8}{3} - \frac{1}{2} + 2 - \frac{1}{3} } $$
$$ \displaystyle { = 8 - \frac{9}{3} - \frac{1}{2} } $$
$$ \displaystyle { = 5 - \frac{1}{2} } $$
$$ \displaystyle { = \frac{10}{2} - \frac{1}{2} } $$
$$ \displaystyle { = \frac{9}{2} } $$
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