Given is the following double integral - $$\iint\limits_{D}\frac{1}{(x^2+ y^2)^2}\,dy\,dx$$
$D$ is the region for $x\in[1,2]$ and $y\in[0,\sqrt{2x-x^2}]$
How would the limits of integration be derived if one was to convert the coordinate system from Cartesian to polar?
So far, I can only understand that $\theta$ would hold an upper bound of $\frac{\pi}{4}$ with a lower bound of $0$ and could derive by algebra that $r$ would hold an upper bound of $2\cos\theta$. By the same algebra, the lower bound for $r$ turns out as $0$.
Here's my go at finding bounds for $r$ $$y = \sqrt{2x - x^2}$$ $$x^2 + y^2 = 2x$$ $$r^2(\cos^2\theta + \sin^2\theta) = 2r\cos\theta$$ $$r^2 - 2r\cos\theta = 0$$ Thus, $r$ is found to be $0, 2\cos\theta$
From a solution that I found, the lower bound for $r$ was given as $\sec\theta$. Could someone please explain how that came to be? I've been at it for a day now
Everything is correct except for the lower bound which is given by
$$x=1\implies r=\sec\theta$$
The integral in polar coordinates is
$$\int_{0}^{\frac{\pi}{4}}\int_{\sec\theta}^{2\cos\theta} \frac{1}{r^3}\:drd\theta$$