Convert $\int_0^1dx \int_0^{x^2}f(x,y) dy $ to polar integration

Question

Converting $\int_0^1dx \int_0^{x^2}f(x,y) dy $ to polar coordinates:

$(r \cos\theta)^2 = r \sin\theta$, so $ r= tg\theta\sec\theta$, then the result is, $$\int_0^{\frac\pi4} d\theta \int_0^{tg\theta\sec\theta} f(r \cos\theta, r \sin\theta) rdr $$

please teach me how to do it correctly.

Answer 1

The first problem is that you don't have $y=x^2$, that is just the boundary of the region of integration, so saying $(r \cos \theta)^2=r \sin \theta$ is not correct. The other is that the region of integration is beyond the parabola (draw a picture), so $r$ should range from the solution to $x^2=x \sin \theta$ to the edge of the square which is at $r=\frac 1{\cos \theta}$ It should be $$\int_0^{\frac\pi4} d\theta \int_{\sin \theta}^{\frac 1{\cos \theta}} f(r \cos\theta, r \sin\theta) rdr$$

Answer 2

The bounds for the polar integral in the question are for $0\le r\le\sec(\theta)\tan(\theta)$:

However, Koro's comment to Ross Millikan's answer suggests $\sec(\theta)\tan(\theta)\le r\le\sec(\theta)$:

which seems like the region you really want, given limits of the Cartesian integral.