Surface integral in Spherical Polar

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SUMMARY

The discussion centers on evaluating the surface integral of the dot product between the vector field A and the differential area element dS in spherical polar coordinates. The user initially asserts that the integral equals zero due to the orthogonality of the polar direction θ and the radial direction. However, a correction is provided, clarifying that dS is not aligned with the radial direction but is instead defined by a specific vector differential of area. The correct expression for dS is given as (a^2cos(θ)sin²(φ)𝑖 + a²sin(θ)sin²(φ)𝑗 + a²sin(φ)cos(φ)𝑘)dθdφ, which does not simplify to a radial vector.

PREREQUISITES
  • Understanding of spherical polar coordinates
  • Familiarity with vector calculus and surface integrals
  • Knowledge of dot products and orthogonality in vector fields
  • Ability to manipulate vector differential area elements
NEXT STEPS
  • Study the derivation of the vector differential area element in spherical coordinates
  • Learn about the application of the Divergence Theorem in surface integrals
  • Explore examples of surface integrals involving vector fields
  • Investigate the properties of orthogonal vectors in three-dimensional space
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Students and professionals in physics and engineering, particularly those focusing on vector calculus, electromagnetism, and fluid dynamics, will benefit from this discussion.

samjohnny
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Homework Statement


Attached.

Homework Equations



cd7a7dabc3ed67196d363a39be943c45.png

01218db9301c3665178919b26218da4b.png

d9beecbcecba1ca83ecd273a33330938.png

1f5ce7228bc649add6fcd17c1ddbf174.png


The Attempt at a Solution


Hi,

Ok, so for the first part of this question it asks to evaluate the integral of the dot product between A and dS. The magnitude of dS is as shown above, and it is in the radial direction in spherical polar coordinates. Here, A(r) is the unit vector of θ in spherical polar. And since we know that the polar direction θ and the radial direction are orthogonal to each, their dot product must = 0. Therefore, the integral of A.dS must be equal to zero. Is this line of thought correct?

Thanks.
 

Attachments

  • Spherical.JPG
    Spherical.JPG
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samjohnny said:

Homework Statement


Attached.

Homework Equations



cd7a7dabc3ed67196d363a39be943c45.png

01218db9301c3665178919b26218da4b.png

d9beecbcecba1ca83ecd273a33330938.png

1f5ce7228bc649add6fcd17c1ddbf174.png


The Attempt at a Solution


Hi,

Ok, so for the first part of this question it asks to evaluate the integral of the dot product between A and dS. The magnitude of dS is as shown above, and it is in the radial direction in spherical polar coordinates. Here, A(r) is the unit vector of θ in spherical polar. And since we know that the polar direction θ and the radial direction are orthogonal to each, their dot product must = 0. Therefore, the integral of A.dS must be equal to zero. Is this line of thought correct?

Thanks.
No, the direction of d\underline{S} is not "in the radial direction". The "vector differential of area" is (a^2cos(\theta)sin^2(\phi)\vec{i}+ a^2sin(\theta)sin^2(\phi)\vec{j}+ a^2 sin(\phi)cos(\phi)\vec{k})d\theta d\phi. That is not a multiple of a radial vector which would be of the form r cos(\theta)sin(\phi)\vec{i}+ r sin(\theta)sin(\phi)\vec{j}+ r cos(\phi)\vec{k}.
 
HallsofIvy said:
No, the direction of d\underline{S} is not "in the radial direction". The "vector differential of area" is (a^2cos(\theta)sin^2(\phi)\vec{i}+ a^2sin(\theta)sin^2(\phi)\vec{j}+ a^2 sin(\phi)cos(\phi)\vec{k})d\theta d\phi. That is not a multiple of a radial vector which would be of the form r cos(\theta)sin(\phi)\vec{i}+ r sin(\theta)sin(\phi)\vec{j}+ r cos(\phi)\vec{k}.

Thanks for the reply. But if I'm thinking about this right, couldn't you factor out of that vector the scalar a^2*sin(α) which would leave the unit vector in the radial direction? Please see the attached.
 

Attachments

  • Notes.JPG
    Notes.JPG
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