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anonymous

  • 4 years ago

let S be the part of the surface z=4-x^2-y^2 that is above z=0. then the value of ln(doubleintegral:(z+2)dS) is:

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  1. anonymous
    • 4 years ago
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    i think you need to use spherical coordinates,

  2. anonymous
    • 4 years ago
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    help.

  3. dumbcow
    • 4 years ago
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    ok i think i can set it up for you: to determine the bounds of the integral use the inequality: 4-x^2 -y^2 > 0 y^2 < 4-x^2 -sqrt(4-x^2) < y < sqrt(4-x^2) Domain inside radical must be positive: 4-x^2 > 0 x^2 < 4 -2 < x < 2 \[\large \int\limits_{-2}^{2}\int\limits_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} (6 -x^{2} -y^{2}) dy dx\]

  4. anonymous
    • 4 years ago
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    Actually, That may just help!, can i go 4* the integrand and go from 0-->sqrt{4-x^2}and 0--->2

  5. anonymous
    • 4 years ago
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    ..instead?

  6. anonymous
    • 4 years ago
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    by symmetry? i'm just trying to save myself a tiny bit of work, but i guess i can just try crunch it out

  7. dumbcow
    • 4 years ago
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    yep...good idea

  8. anonymous
    • 4 years ago
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    Darn, that still didn't work.

  9. dumbcow
    • 4 years ago
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    what didn't work...not the right answer? i get 16pi...so ln(16pi) = 3.917 ??

  10. anonymous
    • 4 years ago
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    yeah, that's what I got as well I'm given 8 possible decimal numbers as a solution and unfortunately that isn't one of them

  11. dumbcow
    • 4 years ago
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    oh ok sorry i checked it again and it seems correct to me but im not an expert when it comes to surface integrals

  12. anonymous
    • 4 years ago
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    Yes and it makes sense to me too, however we must be missing something here. If only that were clear. Thank you much for your efforts

  13. anonymous
    • 4 years ago
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    ln(132.598)=4.88732 This IS one of the answers which means it is correct because all the others are there to prevent guessing! I've seen Paul's notes before, and although they are helpful I was never able to put the full solution together! Thank you alot!

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