Use Stokes' Theorem to Calculate F on Triangle 1,0,0...0,1,0...0,0,1

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SUMMARY

This discussion focuses on applying Stokes' Theorem to calculate the vector field F = xyi + yzj + zxk over the triangular surface defined by the vertices (1,0,0), (0,1,0), and (0,0,1). The curl of F is computed as curl F = -yi - zj - xk, leading to the evaluation of integrals over the defined surface. The solution involves determining the equation of the plane z = 1 - x - y and calculating the gradient, resulting in a series of integrals that yield the final result of 2 - z/2.

PREREQUISITES
  • Understanding of Stokes' Theorem
  • Knowledge of vector calculus, specifically curl and gradient
  • Familiarity with surface integrals and double integrals
  • Ability to work with triangular surfaces in three-dimensional space
NEXT STEPS
  • Study the application of Stokes' Theorem in different geometrical contexts
  • Learn about calculating curl and divergence in vector fields
  • Explore surface integrals and their applications in physics
  • Investigate the properties of gradients and their geometric interpretations
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Students and professionals in mathematics, physics, and engineering who are working with vector calculus and need to apply Stokes' Theorem for surface integrals.

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Homework Statement



Use stokes theorem

F = xyi + yzj + zxk on triangle 1,0,0,,,,,,,0,1,0,,,,,,0,0,1


Homework Equations





The Attempt at a Solution



First i found the curl F

curl F = -yi - zj - xk

Then i found the equation of the plane for the triangle

z = g(xy) = 1 - x - y

Then i found the gradient

f(xyz) = z - g(xy) = z - 1 + x + y, grad f = i + j + k

so the integral will be dy from 0 to x-1, and dx from 0 to 1

(-yi - zj - xk) dot (i + j + k) = -y - z - x

integral (-y - z - x)dy = -y2/2 - zy - xy evaluated from 0 to x-1

integral -(x-1)2/2 - z(x-1) - x(x-1) dx

integral -x2/2 + x - 1/2 - zx + z - x2 + x dx evaluated from 0 to 2

-1/2 + 1 - z/2 + z - 1/2 = 2 - z/2 should that z be in there
 
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[tex] \oint\limits_C {{\rm \bar F \bullet }d{\rm \bar r} = \int\limits_S {\int {(\bar \nabla \times } } } {\rm \bar F}) \bullet {\rm \bar n }\;d{\rm A}[/tex]


Since the surface S can be any surface, any convenient surface can be chosen. This is best done here by choosing the three triangles formed by the three positive axes and the lines joining the three given points.

Consider one such triangle -- for example, the one lying on the x-y plane, which is bounded by the x axis, y-axis and the segment joining (1,0,0) and (0,1,0). The unit normal on this surface is constant and the double integral can be found very easily, taking the proper limits for x and y. The other two can be done similarly (or just by inspection using symmetry property) and all the three then added to get the result.
 

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