Switch to polar coordinates and Then change order of integral $\int_{0}^{1} \int_{0}^{x^2} f(x,y) dy dx$ Substitute to polar coordinates and change order of integral 

$$\int_{0}^{1} \int_{0}^{x^2} f(x,y) dy dx$$

I could substitute to polar coordinates, but failed to change the order of integral
$x=r\cos\theta$, $y=r\sin\theta$.
$$\int_{0}^{1} \int_{0}^{x^2} f(x,y) dy dx = \int_{0}^{\frac{\pi}{4}}d\theta \int_{\frac{\sin\theta}{{\cos^2\theta}}}^{\frac{1}{\cos\theta}} r f(r\cos\theta,r\sin\theta) dr$$.
ADDED
$r=\frac{1}{cos\theta}$, $\theta = \arccos\frac{1}{r}$

$r=\frac{\sin\theta}{{\cos^2\theta}}$, $\sin\theta= r(1-sin^2\theta)$, $sin\theta = \frac{-1\pm \sqrt{1+4r^2}}{2r}$, $\theta = \arcsin \frac{-1\pm \sqrt{1+4r^2}}{2r}$
when $\theta = 0$, $r=0$ to $r=1$, when $\theta = \frac{\pi}{4}$, $r=\sqrt{2}$

My problem is, that when $r \rightarrow 0$, I am not able to see what is happening. I need $r$, $\theta$ dependency graph.
 A: The area you are integrating over is shown in blue in the figure below.

In the integral, you are first varying $y$ from $0$ to $x^2$ and varying $x$ from $0$ to $1$, i.e., you are integrating the red strips and move the red strips from left to right. Instead, we could also proceed as follows. Vary $x$ from $\sqrt{y}$ to $1$ and vary $y$ from $0$ to $1$, i.e., you are integrating the green strips and move the green strips from bottom to top. Hence, we have
$$\int_0^1 \int_0^{x^2} f(x,y) dy dx = \int_0^1 \int_{\sqrt{y}}^1 f(x,y) dx dy$$
A: I don't think polar coordinates are useful for this problem.
Note you are integrating over the region $\{(x,y): 0\leq x \leq 1, 0\leq y \leq x^2\}$. That is, you are integrating over the region in the plane between $x=0$ and $x=1$ under the graph of the function $y = x^2$. Try to make a sketch of the region.
Then you will find that the region can also be expressed as the region where $0\leq y \leq 1$ and $? \leq x \leq 1$. Find the $?$.
A: To see the the angle bound first, think of an circular arc sweeping the area under the question.  When you switch it around, you have to split into 2 integrals, broken at the circle $r=1$. 
For $r<1$, you uniformly have $\theta=0$ for the lower bound.  For the upper bound you only hit the curve $y=x^2$.  you get
$$\int_0^{1} \int_0^{(y=x^2)} f(r,\theta)rd\theta dr$$
For the $y=x^2$ bound you have to convert to polar and solve for $\theta$, like you did, which gives you the ugly arcsin expression you have.
$$\int_0^1 \int_0^{\arcsin \frac{-1\pm \sqrt{1+4r^2}}{2r}} f(r,\theta)rd\theta dr$$
For $r>1$ the lower bound on $\theta$ becomes the vertical line $x=1$.  So you obtain
$$\int_1^\sqrt{2} \int_{csc \theta}^{\arcsin \frac{-1\pm \sqrt{1+4r^2}}{2r}} f(r,\theta)rd\theta dr$$
