Double integral - Convert to polar coordinates and find the integration limits by a given domain I need help converting to polar coordinates and find the limits of the integrals by this given domain:
$$\iint_{D}{} f(x,y)\, dx\, dy$$
$$D= \left\{ (x,y) \mid \dfrac {x^2}{a} \leq y\leq a, -a\leq 0 \leq a   \right\}$$
The solution is: 
 A: I will assume that your third line $a<0$ is an error and that your second line $-a\le 0\le a$ is correct. That second line restriction simplifies to just $0\le a$.
You want to integrate the function over the blue region in this graph. In this particular graph, $a=2$, but everything other than the axes values are true for any positive value of $a$.

Your restriction $\frac{x^2}a\le y\le a$ makes the region above the parabola $y=\frac{x^2}a$ and below the horizontal line $y=a$. The "corners" of that region are at the parabola vertex $(0,0)$ and the intersection points $(\pm a,a)$ of the two graphs.
For any point in Area 1 we can see that $\theta$, the standard angle at the origin, is between $\frac{\pi}4$ and $\frac{3\pi}4$. (Geogebra, my graphing program, only does angle values in degrees, as far as I know.) Note that these angles do not depend on the value of $a$. We also see that $r$, the distance of the point from the origin, has a minimum of $0$ and a maximum of a point on the line $y=a$. Changing that to polar coordinates, the maximum $r$ is determined by
$$r\sin\theta=a$$
$$r=\frac{a}{\sin\theta}$$
We can calculate the area of a region using polar coordinates by
$$\iint_D f(x,y)\,dA=\int_a^b \int_0^{r(\theta)} f(r\cos\theta,r\sin\theta)r\,dr\,d\theta$$
Combining all that, we get the value of Area 1 as
$$\int_{\pi/4}^{3\pi/4} \int_0^{a/\sin\theta} f(r\cos\theta,r\sin\theta)r\,dr\,d\theta$$
That,  of course, is the first integral in your given answer.
For Area 2, we see that $\theta$ goes from $0$ to $\frac{\pi}4$, and the upper limit on $r$ is given by a point on the parabola $\frac{x^2}a=y$,
$$\frac{(r\cos\theta)^2}a=r\sin\theta$$
$$r=\frac{a\sin\theta}{\cos^2 \theta}$$
Using that gives the second integral in your given answer. Area 3 is the same except that $\theta$ goes from $\frac{3\pi}4$ to $\pi$, yielding the third integral in your given answer.
The final answer combines Areas 1, 2, and 3, by summing those three integrals.
