Question

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:

Answer 1

i think you need to use spherical coordinates,

Answer 2

help.

Answer 3

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

\[\large \int\limits_{-2}^{2}\int\limits_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} (6 -x^{2} -y^{2}) dy dx\]

Answer 4

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

Answer 5

..instead?

Answer 6

by symmetry? i'm just trying to save myself a tiny bit of work, but i guess i can just try crunch it out

Answer 7

yep...good idea

Answer 8

Darn, that still didn't work.

Answer 9

what didn't work...not the right answer?

i get 16pi...so ln(16pi) = 3.917 ??

Answer 10

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

Answer 11

oh ok sorry i checked it again and it seems correct to me but im not an expert when it comes to surface integrals

Answer 12

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

Answer 13

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

Answer 14

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!