Given the region $$ R:(x, y | x^2 + y^2 \le 4x) $$ And given the function $f(x, y) = \frac{x}{\sqrt{x^2+y^2}}$, find the double integral of the polar region.
So upon sketching the graph we get a circle with a radius of $2$ centered at $(2,0)$. Converting the given eqn. of $f(x, y)$ into polar coordinates we get that $r^2 \le 4x$, which can be converted to $r^2 \le 4rcos\theta$, and then simplified to $r = 4cos\theta$.
Okay sounds good right? I currently have my region set up in the following manner,
$$D:{(r,\theta)|0\le r\le 4cos\theta, 0 \le \theta \le }$$
however, I'm confused on how to find the upper bound for theta. Without this I can't properly evaluate the double integral, can anyone help me in figuring out how to find this upper bound for theta and why it is so? It cannot be $\frac{\pi}{2}$ because that only encompasses the top half of the circle, the region consists of the entire circle, whose the other half lies in the 4th quadrant.
I will provide a picture of my current work below.
Thank you!
If you use the polar coordinates $$x=r\cos(\theta)\;\;,\;\; y=\sin(\theta)$$ then, because the circle with $ (2,0) $ as center and $ R=2$ as radius, is tangent to the $y-$ axis, at the origine, the angle $ \theta $ will variate between $ -\frac \pi2 $ and $ +\frac \pi2 $.
for each $ \theta $, we have $$0\le r \le R_{max} $$ with $$\cos(\theta)=\frac{2+2\cos(2\theta)}{R_{max}}$$ thus
$$R_{max}=4\cos(\theta)$$ So, $$D=\{(r,\theta)| -\frac \pi2\le \theta\le \frac \pi2 \text{ and } 0\le r\le 4\cos(\theta)\}$$