Help doing an integral using stokes theorem?

Click For Summary
SUMMARY

The discussion centers on applying Stokes' Theorem to evaluate a surface integral involving the vector field F = xi + x³y²j + zk over the boundary of a semi-ellipsoid. The curl of F was calculated to be curl F = 3x²y²k, and the integral was transformed into a double integral over the region defined by the semi-ellipsoid. The user successfully evaluated the integral using Wolfram Alpha, obtaining the result of π, but sought assistance in converting the integral into polar coordinates for manual evaluation. The suggested substitution y = 2y₁ simplifies the integral to ∫∫24x²y₁² dx dy₁ over the unit disk in the (x, y₁) plane.

PREREQUISITES
  • Understanding of Stokes' Theorem and its application in vector calculus.
  • Familiarity with calculating curl of vector fields.
  • Knowledge of double integrals and their evaluation techniques.
  • Proficiency in converting Cartesian coordinates to polar coordinates.
NEXT STEPS
  • Learn how to apply Stokes' Theorem in various contexts beyond simple examples.
  • Study the process of calculating the curl of vector fields in three dimensions.
  • Explore techniques for evaluating double integrals, particularly in polar coordinates.
  • Practice converting integrals from Cartesian to polar coordinates with different boundary conditions.
USEFUL FOR

Students and educators in calculus, particularly those focusing on vector calculus, as well as anyone looking to deepen their understanding of Stokes' Theorem and integral evaluation techniques.

madcattle
Messages
10
Reaction score
0

Homework Statement


F= xi + x3y2j + zk
C is the boundary of the semi-ellipsoid z=√(4-4x2-y2) in the plane z=0


Homework Equations



Stokes theorem states:
∫∫(curlF ° n)dS

The Attempt at a Solution


I found the curl of the F to be 3x2y2k
I found that the dot product of CurlF and n = 3x2y2 divided by dS

Then,
∫∫(3x2y2)/(dS)*dS
=∫∫(3x2y2)dxdy
I evaluated this integral on Wolfram Alpha with the following boundaries:
y goes from -√(4-4x2) to √(4-4x2)
and x from -1 to 1 and got the correct answer = ∏

However, I am finding it impossible (for me!) to do the integral by hand and am wondering if someone can help me turn this into polar coordinates or something else that makes it more solvable

Thank!
 
Physics news on Phys.org
You want to integrate on the region [itex]x^2 + \left( \frac{y}{2} \right)^2 \leq 1[/itex]. Consider the substitution [itex]y = 2y_1[/itex]. Then your integral becomes:

[tex]\int_{D} 24 x^2 y_1^2\ dx dy_1[/tex]

Where D is now the unit disk in the (x, y_1) plane centered at the origin. Can you see how to put that into polar coordinates?
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
6
Views
2K
Replies
6
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K