Multiple integral Issue I'm given the following exercise:
$\iint\limits_D \exp(x^{2}+y^{2})dA$
And I dont even know where to start, any chance someone could give me a hint?
D is a half circle, given by:
$9\le x^{2}+y^{2}\le 16, y\ge 0$
Wheres A, I don't know what it is, its not given in the exercise. - Thats partly whats bugging me.
 A: 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 $r^{2}=x^{2}+y^{2}$:
$$\begin{eqnarray*}
\underset{D}{\iint }e^{x^{2}+y^{2}}\mathrm{d}A &=&\underset{D}{\iint }
e^{x^{2}+y^{2}}\mathrm{d}x\mathrm{d}y \\
&=&\int_{0}^{\pi }\left( \int_{3}^{4}e^{r^{2}}rdr\right) \mathrm{d}\theta  \\
&=&\int_{0}^{\pi }\left[ \frac{1}{2}e^{r^{2}}\right] _{3}^{4}\mathrm{d}\theta  \\
&=&\int_{0}^{\pi }\frac{1}{2}(e^{16}-e^{9})\mathrm{d}\theta  \\
&=&\frac{\pi }{2}(e^{16}-e^{9}).
\end{eqnarray*}$$
A: The A in $dA$ refers to the infinitesimal thing you're integrating over. It just means that instead of a line, you integrate over a surface. A for Area.
As for a hint to start: Since you are dealing with circles, polar coordinates are a natural choice. I am sure these coordinates have been discussed in your lecture before. Note that $x^2 + y^2 = r^2$, where $r$ is the distance of the point from the coordinate origin. Also note that $dA = r dr d\phi$
Then, you can characterize the area $D$ by simple intervals for your polar coordinates: r will go from $3$ to $4$ and $\phi$ will go from $0$ to $\pi$.
A: Some confusion is coming from not understanding what the $dA$ represents, so here's a link on the area element in polar coordinates that hopefully will help clear up this point.
