SOLUTION 18: Compute the area of the enclosed region bounded by the graphs of the equations $x=y+3$ and $x=y^{2}-y$ . Begin by finding the points of intersection of the two graphs. From $x=y+3$ and $x=y^{2}-y$ we get that $$y+3 = y^{2}-y \ \ \longrightarrow$$ $$0 = y^{2} - 2y - 3 \ \ \longrightarrow$$ $$0 = (y-3)(y+1) \ \ \longrightarrow \ \ y=3 \ \ or \ \ y=-1$$ Now see the given graph of the enclosed region.

Using horizontal cross-sections to describe this region, we get that $$-1 \le y \le 3 \ \ and \ \ y^2-y \le x \le y+3 \ ,$$ so that the area of this region is $$AREA = \displaystyle{ \int_{-1}^{3} (Right \ - \ Left) \ dy }$$ $$= \displaystyle { \int_{-1}^{3} ((y+3)-(y^{2}-y)) \ dy }$$ $$= \displaystyle { \int_{-1}^{3} (-y^{2}+2y+3) \ dy }$$ $$= \displaystyle { \Big( -\frac{y^{3}}{3} + y^{2} +3y \Big) \Big\vert_{-1}^{3} }$$ $$= \displaystyle { \Big( -\frac{(3)^{3}}{3} + (3)^{2} + 3(3) \Big) - \Big( -\frac{(-1)^{3}}{3} + (-1)^{2} + 3(-1) \Big) }$$ $$= \displaystyle { (9) - \Big( \frac{1}{3} - 2 \Big) }$$ $$= \displaystyle { \frac{27}{3} - \Big( \frac{1}{3} - \frac{6}{3} \Big) }$$ $$= \displaystyle { \frac{27}{3} - \Big( -\frac{5}{3} \Big) }$$ $$= \displaystyle { \frac{32}{3} }$$