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:

1. anonymous

i think you need to use spherical coordinates,

2. anonymous

help.

3. anonymous

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

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

5. anonymous

6. 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

7. anonymous

yep...good idea

8. anonymous

Darn, that still didn't work.

9. anonymous

what didn't work...not the right answer? i get 16pi...so ln(16pi) = 3.917 ??

10. 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

11. anonymous

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

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
14. 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!