How to Compute F on a Given Surface with Downward Pointing Normal?

lembeh
Messages
4
Reaction score
0

Homework Statement



Let S be the surface given by the graph z = 4 - x2 - y2 above the xy-plane (that it is, where z \geq 0) with downward pointing normal, and let

F (x,y,z) = xcosz i - ycosz j + (x2 + y2 ) k

Compute \oint\oints\oints F dS. (F has a downward pointing normal)

(Hint: Its easy to see that div F = 0 on all R3. This implies that there exists a vector field G such that F = Curl G, although it doesn't tell you what G is)



Homework Equations



z = 4 - x2 - y2 above the xy-plane (that it is, where z \geq 0) with downward pointing normal

F (x,y,z) = xcosz i - ycosz j + (x2 + y2 ) k

Compute \oint\oints\oints F dS. (F has a downward pointing normal)

The Attempt at a Solution



Im getting throw off a bit by the hint. I know its something to do with the surface not being defined around the origin but that's about it.

Homework Statement



See above

Homework Equations



How do I solve this?!
 
Physics news on Phys.org
Looks to me like the hint is suggesting you use Stokes' theorem:
\int \vec{G}\cdot\d\vec{r}= \int\int \nabla G\cdot d\vec{S}
 
Right, but how do I compute this? My daughter hasnt gone past Green's theorem yet in class...I saw this problem on her homework but she couldn't solve it. I can help her and know some Multivariable calculus (but not vector calculus). I want to help her get through this. I would really appreciate it if someone spelt out the solution for me. So I could learn this and help her out with this. I hope that's not an unreasonable request :)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top