Change the integral int[0 to 2]int[0 to sqrt(1-(x-1)^2)] (x+y)/(x^2+y^2) dx dy into an equivalent polar integral.

Question

anonymous · Answer

$$\int\limits_{0}^{2}\int\limits_{0}^{\sqrt{1-(x-1)^{2}}}(x+y)/(x ^{2}+y ^{2})dy dx$$

anonymous · Answer

O_O

anonymous · Answer

That's actually intimidating looking... any idea @TuringTest ?  ^_^

I know only this to help:
x = r cos $\theta$
y = r sin $\theta$

Substitute somehow?

anonymous · Answer

What the heck is r?  1?  Unit circle identity usage on sin^2 + cos^2 = 1?

turingtest · Answer

well we need to know what the region, which is a circle centered at (1,0)
yes the radius is 1 because the first bounds imply$$0\le y\le\sqrt{1-(x-1)^2}\implies(x-1)^2+y^2=1$$which is the equation of the circle I just described\]

anonymous · Answer

$$x=rcos \theta+1, y=rsin$$

turingtest · Answer

@chanh_chung has the right substitution

anonymous · Answer

So...$$r ^{2}\cos ^{2}\theta -2r \cos \theta +r ^{2}\sin ^{2}\theta =0$$ and solve for r?

anonymous · Answer

0<=r<=1

anonymous · Answer

what?

anonymous · Answer

$$\int\limits_{0}^{1}\int\limits_{0}^{2\pi}.....d \theta dr$$

anonymous · Answer

how do you know r=1?

turingtest · Answer

read my post again please

anonymous · Answer

yes,

anonymous · Answer

Ok I just checked and it says the answer is:
$$\int\limits\limits_{0}^{2}\int\limits\limits_{0}^{2\cos \theta}...dr d \theta$$

experimentx · Answer

it seems it can be done without substitution.

turingtest · Answer

but then why is the integral in terms of r and theta?

turingtest · Answer

I don't see how r=2

experimentx · Answer

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