SOLUTION 17: Compute the area of the region enclosed by the graphs of the equations $x=y^{2}$ and $x=4$ . Begin by finding the points of intersection of the two graphs. From $x=y^{2}$ and $x=4$ we get that $$y^{2} = 4 \ \ \longrightarrow$$ $$y^{2} - 4 = 0 \ \ \longrightarrow$$ $$(y+2)(y-2) = 0 \ \ \longrightarrow \ \ y = -2 \ \ or \ \ y = 2$$ Now see the given graph of the enclosed region.

Using horizontal cross-sections to describe this region, we get that $$-2 \le y \le 2 \ \ and \ \ y^2 \le x \le 4 \ ,$$ so that the area of this region is $$AREA = \displaystyle{ \int_{-2}^{2} (Right \ - \ Left) \ dy }$$ $$= \displaystyle { \int_{-2}^{2} (4 - y^{2}) \ dy }$$ $$= \displaystyle { \Big( 4y - \frac{y^{3}}{3} \Big) \Big\vert_{-2}^{2} }$$ $$= \displaystyle { \Big( 4(2) - \frac{(2)^{3}}{3} \Big) - \Big( 4(-2) - \frac{(-2)^{3}}{3} \Big) }$$ $$= \displaystyle { \Big( 8 - \frac{8}{3} \Big) - \Big( -8 + \frac{8}{3} \Big) }$$ $$= \displaystyle { 16 - \frac{16}{3} }$$ $$= \displaystyle { \frac{48}{3} - \frac{16}{3} }$$ $$= \displaystyle { \frac{32}{3} }$$

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