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anonymous

  • 5 years ago

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.

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  1. anonymous
    • 5 years ago
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    so you want bounds in r and theta or x and y?

  2. anonymous
    • 5 years ago
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    x and y

  3. anonymous
    • 5 years ago
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    I know r and theta is easier but, I want to wrap my head better around setting up the bounds in x and y

  4. anonymous
    • 5 years ago
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    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

  5. anonymous
    • 5 years ago
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    I'd say 0 <= x <=4 and y<= sqrt(4-(x-2)^2) but I dont know what <=y

  6. anonymous
    • 5 years ago
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    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.

  7. anonymous
    • 5 years ago
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    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

  8. anonymous
    • 5 years ago
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    so the dy bound is good then, man, I was confusing the mess out of myself.

  9. anonymous
    • 5 years ago
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    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.

  10. anonymous
    • 5 years ago
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    right on, thanks for your help

  11. anonymous
    • 5 years ago
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    well actually that's u helped me lol. totally forget about changing order

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