Verification Of Stoke's Theorem

  • Thread starter Thread starter Baumer8993
  • Start date Start date
  • Tags Tags
    Theorem
Click For Summary
SUMMARY

This discussion focuses on verifying Stoke's Theorem by computing both the surface integral and the line integral for the vector field F = <2x-y, y*z², y²z> over the surface σ defined by z = sqrt(4-x²-y²) above the xy-plane. The user calculated the curl of F as <0,0,1> and obtained a surface integral result of 33.51. In contrast, the line integral, parametrized by <2cos(t), 2sin(t)>, yielded a result of 12.5. The inconsistency between these results raises questions about the correctness of the calculations.

PREREQUISITES
  • Understanding of vector calculus, specifically Stoke's Theorem.
  • Familiarity with surface integrals and line integrals.
  • Knowledge of curl and normal vectors in three-dimensional space.
  • Ability to perform parametrization of curves in vector fields.
NEXT STEPS
  • Review the derivation of curl for vector fields, specifically for F = <2x-y, y*z², y²z>.
  • Study the process of computing surface integrals over parametrized surfaces.
  • Learn about the parametrization of curves in polar coordinates for line integrals.
  • Practice verifying Stoke's Theorem with different vector fields and surfaces.
USEFUL FOR

Students and educators in advanced calculus or vector calculus courses, particularly those focusing on the application of Stoke's Theorem and integral calculus in three-dimensional space.

Baumer8993
Messages
45
Reaction score
0

Homework Statement


Verify Stoke's Theorem by computing both integrals: (stoke's theorem, and the original work integral).
σ is the portion of z = sqrt(4-x2-y2) above the xy-plane and the vector field is F = <2x-y, y*z2, y2z>

Homework Equations


stoke's theorem, and work integral


The Attempt at a Solution


When I do stoke's theorem I get <0,0,1> for my curl F. My normal vector is <2x, 2y, 2z>. After I do the two, and do the integral I get 33.51 for my answer.

My work integral I have <2cos(t), 2sin(t)> for my parametrization path. However, when I do the integral I get 12.5. I am not sure which one is right since I have tried the problem three times, but I always get the same answer, so I do not know what to do.
 
Physics news on Phys.org
Baumer8993 said:

Homework Statement


Verify Stoke's Theorem by computing both integrals: (stoke's theorem, and the original work integral).
σ is the portion of z = sqrt(4-x2-y2) above the xy-plane and the vector field is F = <2x-y, y*z2, y2z>

Homework Equations


stoke's theorem, and work integral


The Attempt at a Solution


When I do stoke's theorem I get <0,0,1> for my curl F. My normal vector is <2x, 2y, 2z>. After I do the two, and do the integral I get 33.51 for my answer.

My work integral I have <2cos(t), 2sin(t)> for my parametrization path. However, when I do the integral I get 12.5. I am not sure which one is right since I have tried the problem three times, but I always get the same answer, so I do not know what to do.

Do you expect us to work both sides out to see what we get? Show us what you did and we can likely quickly find your mistake.
 

Similar threads

  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
20
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 21 ·
Replies
21
Views
4K
Replies
6
Views
3K
Replies
7
Views
2K