Surface integral in Spherical Polar

In summary, the student is trying to find an equation for the integral of the dot product between the unit vector A and the vector dS in spherical polar coordinates. However, the direction of d\underline{S} is not in the radial direction, so the student needs to factor out the scalar a^2*sin(α) which would leave the unit vector in the radial direction.
  • #1
samjohnny
84
1

Homework Statement


Attached.

Homework Equations



cd7a7dabc3ed67196d363a39be943c45.png

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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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  • #2
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 [itex]d\underline{S}[/itex] is not "in the radial direction". The "vector differential of area" is [itex](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[/itex]. That is not a multiple of a radial vector which would be of the form [itex]r cos(\theta)sin(\phi)\vec{i}+ r sin(\theta)sin(\phi)\vec{j}+ r cos(\phi)\vec{k}[/itex].
 
  • #3
HallsofIvy said:
No, the direction of [itex]d\underline{S}[/itex] is not "in the radial direction". The "vector differential of area" is [itex](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[/itex]. That is not a multiple of a radial vector which would be of the form [itex]r cos(\theta)sin(\phi)\vec{i}+ r sin(\theta)sin(\phi)\vec{j}+ r cos(\phi)\vec{k}[/itex].

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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1. What is a surface integral in Spherical Polar?

A surface integral in Spherical Polar is a mathematical tool used to calculate the flux across a surface in three-dimensional space. It involves integrating a function over a surface that is defined by spherical coordinates (ρ, θ, φ).

2. How is a surface integral in Spherical Polar different from a regular surface integral?

A surface integral in Spherical Polar is different from a regular surface integral because it takes into account the curvature of the surface. Spherical coordinates are used to describe the surface, rather than Cartesian coordinates, which are used in a regular surface integral.

3. What is the formula for calculating a surface integral in Spherical Polar?

The formula for calculating a surface integral in Spherical Polar is given by ∫∫S f(ρ, θ, φ) dS = ∫∫D f(ρ(θ, φ), θ, φ) ρ2 sin(θ) dθ dφ, where S is the surface, D is the region in the ρ-θ-φ coordinate system, and f(ρ, θ, φ) is the function being integrated.

4. What is the significance of the ρ2 sin(θ) term in the surface integral formula?

The ρ2 sin(θ) term in the surface integral formula represents the surface element in Spherical Polar coordinates. It takes into account the radial distance (ρ), the angle θ, and the curvature of the surface (sin(θ)). This term is crucial for accurately calculating the flux across a curved surface.

5. What are some real-world applications of surface integrals in Spherical Polar?

Surface integrals in Spherical Polar have various real-world applications, including calculating the flux of a vector field through a closed surface, determining the electric field around a charged sphere, and finding the mass distribution of a spherical object. They are also used in fluid mechanics, electromagnetism, and quantum mechanics.

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