The notation $\underset{D}{\displaystyle\iint }f(x,y)\;\mathrm{d}A$ has the same meaning as $\underset{D}{\displaystyle\iint }f(x,y)\;\mathrm{d}x\mathrm{d}y$.
To evaluate an integral in a different coordinate system one has to find the absolute value of the Jacobian determinant of the transformation. Using polar coordinates, we can transform the initial integral
$$\underset{D}{ \iint }e^{x^{2}+y^{2}}\;\mathrm{d}A=\underset{D}{\iint }e^{x^{2}+y^{2}}\;\mathrm{d}xdy$$
into this one
$$\int_{0}^{\pi }\left( \int_{3}^{4}e^{r^{2}}rdr\right) \mathrm{d}\theta ,$$
where the conversion factor $r$ is the Jacobian determinant (see Example 3 on the Wikipedia article), and observe that $r^{2}=x^{2}+y^{2}$:
$$\begin{eqnarray*}
\underset{D}{\iint }e^{x^{2}+y^{2}}\mathrm{d}A &=&\underset{D}{\iint }
e^{x^{2}+y^{2}}\mathrm{d}x\mathrm{d}y \\
&=&\int_{0}^{\pi }\left( \int_{3}^{4}e^{r^{2}}rdr\right) \mathrm{d}\theta \\
&=&\int_{0}^{\pi }\left[ \frac{1}{2}e^{r^{2}}\right] _{3}^{4}\mathrm{d}\theta \\
&=&\int_{0}^{\pi }\frac{1}{2}(e^{16}-e^{9})\mathrm{d}\theta \\
&=&\frac{\pi }{2}(e^{16}-e^{9}).
\end{eqnarray*}$$