SOLUTION 19: Compute the area of the region enclosed by the graphs of the equations $x=y^3$ and $x=y^{2}+2y$ . Begin by finding the points of intersection of the two graphs. From $x=y^3$ and $x=y^2+2y$ we get that $$y^{3} = y^2 + 2y \ \ \longrightarrow$$ $$y^{3} - y^2 - 2y = 0 \ \ \longrightarrow$$ $$y ( y^{2} - y - 2 ) = 0 \ \ \longrightarrow$$ $$y ( y-2 ) ( y+1 ) = 0 \ \ \longrightarrow \ \ y=0 , y=2 , \ \ or \ \ y=-1$$ Now see the given graph of the enclosed region.

Using horizontal cross-sections to describe this region, which is made up of two smaller regions, we get that $$-1 \le y \le 0 \ \ and \ \ y^2+2y \le x \le y^3$$ in addition to $$0 \le y \le 2 \ \ and \ \ y^3 \le x \le y^2+2y \ ,$$ so that the area of this region is $$AREA = \displaystyle{ \int_{-1}^{0} (Right \ - \ Left) \ dy + \int_{0}^{2} (Right \ - \ Left) \ dy }$$ $$= \displaystyle { \int_{-1}^{0} (y^{3} - (y^{2}+2y)) \ dy + \int_{0}^{2} ((y^{2}+2y) - y^{3}) \ dy }$$ $$= \displaystyle { \int_{-1}^{0} (y^{3} - y^{2} - 2y) \ dy + \int_{0}^{2} (y^{2}+2y - y^{3}) \ dy }$$ $$= \displaystyle { \Big( \frac{y^{4}}{4} - \frac{y^{3}}{3} - y^{2} \Big) \Big\vert_{-1}^{0} + \Big( \frac{y^{3}}{3} + y^{2} - \frac{y^{4}}{4} \Big) \Big\vert_{0}^{2} }$$ $$= \displaystyle { \Big( \frac{(0)^{4}}{4} - \frac{(0)^{3}}{3} - (0)^{2} \Big) - \Big( \frac{(-1)^{4}}{4} - \frac{(-1)^{3}}{3} - (-1)^{2} \Big) + \Big( \frac{(2)^{3}}{3} + (2)^{2} - \frac{(2)^{4}}{4} \Big) - \Big( \frac{(0)^{3}}{3} + (0)^{2} - \frac{(0)^{4}}{4} \Big) }$$ $$= \displaystyle { \Big( 0 \Big) - \Big( \frac{1}{4} + \frac{1}{3} - 1 \Big) + \Big( \frac{8}{3} + 4 - 4 \Big) - \Big( 0 \Big) }$$ $$= \displaystyle { - \Big( \frac{3}{12} + \frac{4}{12} - \frac{12}{12} \Big) + \frac{8}{3} }$$ $$= \displaystyle { -\Big( -\frac{5}{12} \Big) + \frac{8}{3} }$$ $$= \displaystyle { \frac{5}{12} + \frac{32}{12} }$$ $$= \displaystyle { \frac{37}{12} }$$