## anonymous 5 years ago Double integral 2y dA, where R is region in 1st quadrant bounded above by circle (x-1)^2 + y^2=1 and below by line y=x. This even stumped my teacher.

1. bobbyleary

Do you know the answer to this? I solved it but wanted to see if you found and answer.

2. bobbyleary

$\int\limits \int\limits 2ydA$ $(x-1)^2 + y^2 = 1$$y^2 = 1 - (x-1)^2$$y = \pm \sqrt{1-(x-1)^2}$But since it's only the first quadrant, we are only interested in the positive answer. And we know y = x so we now have our bounding limits for y and x goes from 0 to 1. $\int\limits_{0}^{1}\int\limits_{x}^{\sqrt{1-(x-1)^2}}2ydydx = 1/3$

3. anonymous

Oh great, I was stumped last semester, and new things kept coming every week. I would look it over in the morning. Thanks.

4. bobbyleary

No problem. I'm interested to know if I'm right. Sorry I couldn't help you before.

5. anonymous

Looking back at my notes, this is easier by using polar coordinates. We have a half circle of radius one, centered (1, 0), x runs from 0 to 2. That circle is cut by line y=x. We consider only what is below that line in the half circle. More...

6. anonymous

Equation$(x-1)^{2}+y ^{2}=1$in polar$r(r -2r \cos \theta)=0$$0\le r \le2\cos \theta$$0\le \theta \le(\pi/4)$

7. anonymous

$\int\limits_{0}^{\pi/4}\int\limits_{0}^{2\cos \theta}2r \sin \theta rdrd \theta$

8. anonymous

9. bobbyleary

I found an error, the problem says that it is bounded by the circle above and y=x below so your angle should be from pi/4 to pi/2 instead of 0 to pi/4. When you solve this, you will get 1/3. If you are still interested, check my work.

10. anonymous

Oh, yes, I remember my tutor worked this for me, and at the time I noted he worked the wrong part of the half circle. It should correspond to what you have.

11. anonymous

Yes, I checked my notes where we re-did it and indeed the answer is 1/3.

12. bobbyleary

Good. Glad I could help, just a month late though. Haha.

13. anonymous

It was a good review. If you don't keep doing it you forget. My tutor happen to be a lecturer. And he is teaching this same course. He invited me to just come and sit it and now I can absorb everything.