Flux through a surface Question

  • Thread starter Thread starter leext101
  • Start date Start date
  • Tags Tags
    Flux Surface
Click For Summary
The discussion revolves around calculating the flux of the vector field F through the surface S defined by z=49-(x²+y²)². The initial attempt at solving the problem resulted in an incorrect flux value of -476. Participants suggest using polar coordinates for parameterization, with limits set from 0 to √7 for r and 0 to 2π for θ, leading to a recalculated answer of -238π. Clarifications are made regarding the use of normal vectors versus tangent vectors in the flux calculation. The conversation emphasizes the importance of correctly setting limits and understanding the orientation of the surface.
leext101
Messages
4
Reaction score
0

Homework Statement


Let S be the part of the surface z=49-(x2+y2)2 above the xy-plane, oriented upward.

Let vector field F= (yz) i +(xz) j + (-17+xy) k

Compute the flux of F through S.


Homework Equations


Flux through surface equation ∫s F(x,y,f(x,y)) dot product (-fx i-fy j + k) dxdy


The Attempt at a Solution


I used the equation to find flux through a surface plugging in F(x,y,(49-(x2+y2)2) for the vector field, I took the dot product. I believe the limits are -sqrt(7)≤x≤sqrt(7) and -sqrt(7)≤y≤sqrt(7). The answer I integrated out was -476 which was incorrect.

I appreciate your time and help!
 
Physics news on Phys.org
Try parameterizing the surface in polar coordinates and then use

∫∫F.n ds = ∫∫F |rrxrθ| dA
 
rock.freak667 said:
Try parameterizing the surface in polar coordinates and then use

∫∫F.n ds = ∫∫F |rrxrθ| dA
The first of these should be a single path integral, not a double integral, shouldn't it? Also I am puzzled by your "F.n". I would have used \vec{F}\cdot d\vec{s} where "d\vec{s}" is the vector tangent to the curve, not normal to it, with length ds.
 
So if I do use polar coordinates would the limits be 0≤r≤sqrt(7), 0≤θ≤2∏ and a normal vector of n= k?
 
using polar coordinates I calculated an answer -238pi
 
If I stick with cartesian coordinates would the limits be -sqrt(7)≤y≤sqrt(7) and
-sqrt(7)≤x≤sqrt≤(7)?

Thanks everyone for posting
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K