Using a Double Integral to Find the Area of a Region

Use a double integral to find the area \(A\) of the region \(R\) in the first quadrant enclosed by the parabola \(y=6x-x^{2}\) and the line \(y=4x-8\).

Solution The graph of the region \(R\) is shown in Figure 20. Notice that \(R\) is neither \(x\)-simple nor \(y\)-simple. It is not \(x\)-simple because of the corner at \((2,0)\); it is not \(y\)-simple because of the corner at \((4,8)\). If we partition \(R\) by drawing a vertical line through the point \((2,0)\), we obtain two \(x\)-simple subregions \(R_{1}\) and \(R_{2}\). Then the area \(A\) of the region \(R\) is \begin{eqnarray*} A&=&\displaystyle\iint\limits_{\kern-3ptR}\,{\it dA}=\displaystyle\iint\limits_{\kern-3ptR_{1}}\,{\it dA}+\displaystyle\iint\limits_{\kern-3ptR_{2}}\,{\it dA}= \int_{0}^{2}\left[ \int_{0}^{6x-x^{2}} {\it dy} \right]\! {\it dx}+\int_{2}^{4}\left[ \int_{4x-8}^{6x-x^{2}} {\it dy}\right]\! {\it dx} \notag \\[4pt] & =&\int_{0}^{2}(6x-x^{2})\,{\it dx}+\int_{2}^{4}(-x^{2}+2x+8)\,{\it dx} \notag \\[4pt] & =&\left[ 3x^{2}-\dfrac{x^{3}}{3}\right] _{0}^{2}+\left[ -\dfrac{x^{3}}{3} +x^{2}+8x\right] _{2}^{4}= \dfrac{56}{3}\hbox{ square units} \end{eqnarray*}