anonymous
  • anonymous
can someone tell me how they would set up the bounds for a double integral above a circle (x-2)^2+y^2=4. Not in polar coords.
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
so you want bounds in r and theta or x and y?
anonymous
  • anonymous
x and y
anonymous
  • anonymous
I know r and theta is easier but, I want to wrap my head better around setting up the bounds in x and y

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anonymous
  • anonymous
normally you dont need to find bounds in x and y from r and theta. since knowing bounds in r and theta is just a method to find double integration that have bounds in x and y. Anyway, in this case, the circle has center (2,0) radius 2, so theta runs from 0 to pi/4 and r from 0 to 4 wait... I think im confused too lol
anonymous
  • anonymous
I'd say 0 <= x <=4 and y<= sqrt(4-(x-2)^2) but I dont know what <=y
anonymous
  • anonymous
It would be easier to integrate taking the x and y terms and changing them over to polar, but I want to keep it all in xand y and I'm confused on how I make sure that I stay in the circle, and not set up something wierd and inaccurate. I thought about 2 *\[\int\limits_{0}^{\sqrt{4-(x-2)^2}} dx\] for dx, but I don't think I can go -2 to 2 for dy. I know I could do y= -root for the circle in stead of o for the lower bound in dx, but I'd rather just multiply the sum I listed by two. I just don't know about the dy limits, though.
anonymous
  • anonymous
yeah that's also what confused me. that 0 is not quite right. It cut the circle in half. I prefer something that would intergrate the full circle so that we dont need to time 2. May need to ask someone about this. But I don't think they will ever ask to find x and y bounds from r, theta oh no, I think that's actually right. I was thinking ∫dy but if you have ∫dx there, that will bound the whole circle, then the bound for y is -2 to 2 is fine, because ∫∫dxdy = ∫∫dydx
anonymous
  • anonymous
so the dy bound is good then, man, I was confusing the mess out of myself.
anonymous
  • anonymous
yeah it's good. When you change ∫∫dxdy to ∫∫dydx, you need to change the bounds for x and y, but the meaning of the total bound remain the same.
anonymous
  • anonymous
right on, thanks for your help
anonymous
  • anonymous
well actually that's u helped me lol. totally forget about changing order

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