Finding a Double Integral over a Region That Is \(x \)-Simple and \(y\)-Simple

Find \(\iint\limits_{\kern-3ptR} e^{y^{2}}\,dA\), where \(R\) is the region shown in Figure 17.

Solution The region \(R\) is both \(x\)-simple and \(y\)-simple.

If we treat \(R\) as \(x\)-simple, then the bottom boundary of \(R\) is \(y=2x\) and the top boundary is \(y=2,\) \(0\leq x\leq 1.\) Then by Fubini’s Theorem, \[ \displaystyle\iint\limits_{\kern-3ptR} e^{y^{2}} dA=\int_{0}^{1}\left[ \int_{2x}^{2}e^{y^{2}} dy \right] dx \]

We cannot find \(\int_{2x}^{2}e^{y^{2}} dy\) using the Fundamental Theorem of Calculus since this integral is not expressible in terms of elementary functions.

So, we consider the region \(R\) as \(y\)-simple. Then the left boundary of \(R\) is \(x=0\) and the right boundary is \(x=\dfrac{y}{2},\) where \(0\leq y\leq 2.\) Then by Fubini’s Theorem, \begin{eqnarray*} \displaystyle\iint\limits_{\kern-3ptR} e^{y^{2}}dA &=&\int_{0}^{2}\left[ \int_{0}^{y/2}e^{y^{2}} dx\right] \! dy=\int_{0}^{2} e^{y^{2}} \big[x\big] _{0}^{y/2}\,dy=\int_{0}^{2}\dfrac{y}{2}e^{y^{2}} dy \notag \\[5pt] &=&\dfrac{1}{2}\int_{0}^{2}ye^{y^{2}} dy=\dfrac{1}{2}\left[ \dfrac{e^{y^{2}} }{2}\right] _{0}^{2}=\dfrac{1}{4}(e^{4}-1) \end{eqnarray*}