Calculating the Surface Integral Using Stokes' Theorem

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SUMMARY

The discussion focuses on calculating the surface integral using Stokes' Theorem, specifically addressing the double integral of the curl of vector field F over surface S. The normal vector to surface S is confirmed as (2/7, 6/7, -3/7), and the curl of F is determined to be -2y. The dot product yields 6y/7, leading to the surface integral over the elliptical disk formed by the intersection of the plane and the cylinder. The area element conversion from dS to dxdy is highlighted as a crucial step in the calculation.

PREREQUISITES
  • Understanding of Stokes' Theorem and its application in vector calculus.
  • Familiarity with curl operations in vector fields.
  • Knowledge of surface integrals and their computation.
  • Ability to perform coordinate transformations, particularly from elliptical to circular coordinates.
NEXT STEPS
  • Study the application of Stokes' Theorem in various vector fields.
  • Learn about calculating curl in three-dimensional space.
  • Explore techniques for converting between different area elements in integrals.
  • Practice solving surface integrals involving elliptical and circular geometries.
USEFUL FOR

Students and educators in advanced calculus, particularly those focusing on vector calculus and surface integrals. This discussion is beneficial for anyone seeking to deepen their understanding of Stokes' Theorem and its practical applications in physics and engineering.

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Homework Statement


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Homework Equations



The double integral at S of the (curl of)F.n.

The Attempt at a Solution



We find the the normal vector to the surface S is (2/7, 6/7, -3/7), right or not? We compute the curl of F which is -2y, right? Then, we calculate the dot product and we get 6y/7. Am I doing it right? If so, can someone tell me what to do next, and if not, can someone tell me how to properly solve it. I'm fairly new to the concept so please be patient with me.
 
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Then you'll perform the SURFACE integral of the dot product 6y/7 over the intersection of the plane and the cylinder, which is an elliptical disk. The integral can be converted to a DOUBLE integral with x and y in the circular disk x^2+y^2=1. The trick is to convert the area element dS of the elliptical to the area element dxdy in the circular disk. Intuitively dS*cos(theta)=dxdy, where theta is the angle of the intersection plane to the x-y plane.
 

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