anonymous
  • anonymous
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:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
i think you need to use spherical coordinates,
anonymous
  • anonymous
help.
dumbcow
  • dumbcow
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 \[\large \int\limits_{-2}^{2}\int\limits_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} (6 -x^{2} -y^{2}) dy dx\]

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anonymous
  • anonymous
Actually, That may just help!, can i go 4* the integrand and go from 0-->sqrt{4-x^2}and 0--->2
anonymous
  • anonymous
..instead?
anonymous
  • anonymous
by symmetry? i'm just trying to save myself a tiny bit of work, but i guess i can just try crunch it out
dumbcow
  • dumbcow
yep...good idea
anonymous
  • anonymous
Darn, that still didn't work.
dumbcow
  • dumbcow
what didn't work...not the right answer? i get 16pi...so ln(16pi) = 3.917 ??
anonymous
  • anonymous
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
dumbcow
  • dumbcow
oh ok sorry i checked it again and it seems correct to me but im not an expert when it comes to surface integrals
anonymous
  • anonymous
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
dumbcow
  • dumbcow
hey i found something try 132.598 http://tutorial.math.lamar.edu/Classes/CalcIII/SurfaceIntegrals.aspx http://www.wolframalpha.com/input/?i=integrate+4*%286-x%5E2+-y%5E2%29sqrt%284x%5E2%2B4y%5E2+%2B1%29+dydx+from+y%3D0+to+sqrt%284-x%5E2%29+and+x%3D0+to+2
anonymous
  • anonymous
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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