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anonymous
 5 years ago
Double integral 2y dA, where R is region in 1st quadrant bounded above by circle (x1)^2 + y^2=1 and below by line y=x. This even stumped my teacher.
anonymous
 5 years ago
Double integral 2y dA, where R is region in 1st quadrant bounded above by circle (x1)^2 + y^2=1 and below by line y=x. This even stumped my teacher.

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bobbyleary
 5 years ago
Best ResponseYou've already chosen the best response.1Do you know the answer to this? I solved it but wanted to see if you found and answer.

bobbyleary
 5 years ago
Best ResponseYou've already chosen the best response.1\[\int\limits \int\limits 2ydA\] \[(x1)^2 + y^2 = 1\]\[y^2 = 1  (x1)^2\]\[y = \pm \sqrt{1(x1)^2}\]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. \[\int\limits_{0}^{1}\int\limits_{x}^{\sqrt{1(x1)^2}}2ydydx = 1/3\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh great, I was stumped last semester, and new things kept coming every week. I would look it over in the morning. Thanks.

bobbyleary
 5 years ago
Best ResponseYou've already chosen the best response.1No problem. I'm interested to know if I'm right. Sorry I couldn't help you before.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Looking 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...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Equation\[(x1)^{2}+y ^{2}=1\]in polar\[r(r 2r \cos \theta)=0\]\[0\le r \le2\cos \theta\]\[0\le \theta \le(\pi/4)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{\pi/4}\int\limits_{0}^{2\cos \theta}2r \sin \theta rdrd \theta\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Integrate that, answer is 1

bobbyleary
 5 years ago
Best ResponseYou've already chosen the best response.1I 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, I checked my notes where we redid it and indeed the answer is 1/3.

bobbyleary
 5 years ago
Best ResponseYou've already chosen the best response.1Good. Glad I could help, just a month late though. Haha.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It 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.
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