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Mathematics 30 Online
OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

O_O

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (turingtest):

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\]

OpenStudy (anonymous):

\[x=rcos \theta+1, y=rsin\]

OpenStudy (turingtest):

@chanh_chung has the right substitution

OpenStudy (anonymous):

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

OpenStudy (anonymous):

0<=r<=1

OpenStudy (anonymous):

what?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

how do you know r=1?

OpenStudy (turingtest):

read my post again please

OpenStudy (anonymous):

yes,

OpenStudy (anonymous):

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

OpenStudy (experimentx):

it seems it can be done without substitution.

OpenStudy (turingtest):

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

OpenStudy (turingtest):

I don't see how r=2

OpenStudy (experimentx):

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