Finding a Double Integral Using Polar Coordinates

Find \(\displaystyle\iint\limits_{\kern-3ptR}e^{x^{2}+y^{2}}\,d\!A\), where \(R\) is the region in the first quadrant inside the circle \(x^{2}+y^{2}=a^{2}\).

Solution In rectangular coordinates, \(\iint \limits_{\kern-3ptR}e^{x^{2}+y^{2}}\,d\!A\) only can be approximated using numerical techniques. But the presence of \(x^{2}+y^{2}\) in the integrand and the fact that the region \(R\) has part of a circle as its boundary suggest using polar coordinates. Figure 28 shows the region \(R.\) We use polar coordinates to obtain \[ \begin{eqnarray*} \displaystyle\iint\limits_{\kern-3ptR}e^{x^{2}+y^{2}}\,d\!A &=&\displaystyle\iint\limits_{\kern-3ptR}e^{r^{2}}r\,dr\,d\theta =\int_{0}^{\pi/2}\int_{0}^{a}e^{r^{2}}r\,dr\,d\theta \underset{\underset{\color{#0066A7}{\hbox{\( \begin{array}{c} u=r^{2}\hbox{,}\ du=2r\, dr\\ r=0\hbox{,}\ \hbox{ then } u=0\hbox{;}\ r=a\hbox{,}\ \hbox{ then } u=a^2\end{array}\)}}}{\color{#0066A7}{{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height9pt depth0pt}}\right.}}}}= \int_{0}^{\pi/2}\int_{0}^{a^{2}}\dfrac{1}{2}e^{u}du\,d\theta \\[-17pt] &=&\dfrac{1}{2} \int_{0}^{\pi /2} \big[e^u\big]_0^{a^2} d\theta = \dfrac{1}{2}\int_{0}^{\pi /2}\big(e^{a^{2}}-1\big)\,d\theta = \left( \dfrac{\pi }{4}\right) \big(e^{a^{2}}-1\big) \end{eqnarray*} \]