Finding area of integration in polar?

[um0123]

Homework Statement

Im having trouble find the area of integration for this integral which i have to convert to polar:

$\int_0^2 \int_0^\sqrt{1-(x-1)^2} \frac{x+y}{x^2 + y^2}$ dydx

Homework Equations

x = rcosθ
y = rsinθ
r = x^2 + y^2

The Attempt at a Solution

i know exactly what to do to the integrand, i just don't understand how to turn the upper limit sqrt(1-(x-1)^2) into polar, i can't visualize it at all!

i know it eventually turns into 2cosθ but i don't understand how to get there.

 

[DryRun]

What is the whole question? Have you been given that double integral only? If you deduced those limits on your own, maybe you have them wrong.

 

[LCKurtz]

$y=\sqrt{(1-(x-1)^2}$ is the top half of the circle $(x-1)^2+y^2=1$. Just convert that to polar coordinates and simplify it and be surprised.

 

[DryRun]

$y=\sqrt{(1-(x-1)^2}$ turns into $2\cos \theta$.

To help you to visualize the problem, convert the Cartesian coordinate integral into its polar form (i haven't included the integrand for simplicity, but you will obviously have to convert the latter as well):
$$\int^{\theta=\frac{\pi}{2}}_{\theta=0} \int^{r=2\cos \theta}_{r=0} rdrd\theta$$
Hence, you will realize that you need to express r in terms of $\theta$, which should simplify the polar conversion process.

 

[LCKurtz]

$y=\sqrt{(1-(x-1)^2}$ actually turns into $r=\sqrt{2\cos \theta}$ and not $2\cos \theta$.

No it doesn't. Check your arithmetic.

 

[DryRun]

No it doesn't. Check your arithmetic.

Yes, you are correct. I had substituted r=1 somewhere in the conversion.
I've edited my previous post as well.

 

[LCKurtz]

Yes, you are correct. I had substituted r=1 somewhere in the conversion.
I've edited my previous post as well.

Just a reminder from the rules:

NOTE: ... Once your question or problem has been responded to, do not go back and delete (or edit) your original post.