Question

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.

Answer 1

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

Answer 2

O_O

Answer 3

That's actually intimidating looking... any idea @TuringTest ? ^_^

I know only this to help:
x = r cos \(\theta\)
y = r sin \(\theta\)

Substitute somehow?

Answer 4

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

Answer 5

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\]

Answer 6

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

Answer 7

@chanh_chung has the right substitution

Answer 8

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

Answer 9

0<=r<=1

Answer 10

what?

Answer 11

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

Answer 12

how do you know r=1?

Answer 13

read my post again please

Answer 14

yes,

Answer 15

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

Answer 16

it seems it can be done without substitution.

Answer 17

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

Answer 18

I don't see how r=2