Area of circle (double integral and cartesian coordinates)? I know that the area of a circle, $x^2+y^2=a^2$, in cylindrical coordinates is
$$
\int\limits_{0}^{2\pi} \int\limits_{0}^{a} r \, dr \, d\theta = \pi a^2
$$
But how can find the same result with a double integral and only cartesian coordinates?

Just as an alternative solution to the qbert answer.
Note that $|x|$, $|y|$ are not strictly less than $r$, but instead $|x| \leq r$, $|y| \leq r$. This can still insure that $r^2 \geq y^2$ and so $\sqrt{r^2-y^2}$ is real.
After the first integral
$$
\int_{-r}^r\int_{-\sqrt{r^2-y^2}}^{\sqrt{r^2-y^2}}\mathrm dx\mathrm dy=
\int_{-r}^r2\sqrt{r^2-y^2}\mathrm dy
$$
Instead of the change of variable, you can remember the known integral:
$$\int \sqrt{r^2 - y^2} = \frac{1}{2} \left( r^2 \arcsin \frac{y}{r} + y \sqrt{r^2 - y^2} \right) + C$$
whose condition $|y| \leq |r|$ has just been mentioned and it is satisfied. The result follows almost immediately:
$$\int_{-r}^r 2 \sqrt{r^2 - y^2} \mathrm{d} y = \left[ r^2 \arcsin \frac{y}{r} + y \sqrt{r^2 - y^2} \ \right]_{-r}^r = r^2 \arcsin (1) - r^2 \arcsin (-1) = \\
= r^2 \frac{\pi}{2} + r^2 \frac{\pi}{2} = \pi r^2$$
So, using the cartesian coordinates, the only important observation is simply to consider the right extreme values for each variable. 
Using the circumference equation $x^2 + y^2 = r^2$, you can choose $x$ as a function of $y$, obtaining $x = \pm \sqrt{r^2 - y^2}$, which will be the extreme values for $x$. Then, you let $y$ sweep from $-r$ to $r$, which are the extreme values for $y$. Of course, you can alternatively do vice-versa, with $y$ as function of $x$. 
Unlike the cylindrical coordinates, there is no angular variation here, but only an horizontal variation between $- \sqrt{r^2 - y^2}$ and $\sqrt{r^2 - y^2}$ as regards $x$, and a vertical variation between $-r$ and $r$ as regards $y$.
