Finding Area by Partitioning the \(y\)-Axis

Find the area \(A\) of the region enclosed by the graphs of \(x=f(y)=y+2\) and \(x=g(y)=y^{2}\).

Solution First we graph the two functions and identify the region whose area \(A\) we seek. See Figure 10.

The graphs intersect when \(y+2=y^{2}\). Then \(y^{2}-y-2=(y-2) \left(y+1\right) =0\). So, \(y=2\) or \(y=-1\). When \(y=2, x=4\); when \(y=-1, x=1\). The graphs intersect at the points \((4,2)\) and \(\left(1,-1\right)\).

Notice in Figure 10 that the graph of \(f\) is to the right of the graph of \(g\); that is, \(f(y)\geq g(y)\) for \(-1\leq y\leq 2\). This indicates that we can partition the \(y\)-axis and form rectangles from the left graph (\(x=y^{2}\)) to the right graph \((x=y+2)\) as \(y\) varies from \(-1\) to \(2\). The area \(A\) of the region between the graphs is \begin{eqnarray*} A&=&\int_{-1}^{2}[ f(y)-g(y)] {\it dy}=\int_{-1}^{2}[ ( y+2) -y^{2}] {\it dy}=\left[ \dfrac{y^{2} }{2}+2y-\dfrac{y^{3}}{3}\right] _{-1}^{2}\\[6pt] &=&\left( 2+4-\dfrac{8}{3}\right) -\left( \dfrac{1}{2}-2+\dfrac{1}{3}\right) =4.5\hbox{ square units} \end{eqnarray*}