Surface integral in Spherical Polar

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samjohnny
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Homework Statement


Attached.

Homework Equations



Equation for the spherical polar unit vector r-hat: sin(θ)cos(φ) i-hat + sin(θ)sin(φ) j-hat + cos(θ) k-hat.

Equation for the unit vector in the theta direction in spherical coordinates: theta-hat = cos(theta)cos(phi) i + cos(theta)sin(phi) j - sin(theta) k.

Unit vector in the polar direction: phi-hat = -sin(phi) i + cos(phi) j, written with hat-marked unit vector symbols.

Vector differential area element in spherical polar coordinates: dS_r = r^2 sin θ dθ dφ.


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
  • Typeset problem statement: a spherical surface S of radius a with parametric form r = a sin θ cos φ e1 + a sin θ sin φ e2 + a cos θ e3, 0 ≤ θ ≤ π, 0 ≤ φ ≤ 2π; asks for the area element and surface integrals of A·dS and A×dS over the octant, with A = cos θ cos φ e1 + cos θ sin φ e2 − sin θ e3.
    Spherical.JPG
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samjohnny said:

Homework Statement


Attached.

Homework Equations



Unit radial vector in spherical polar coordinates: r-hat = sin(theta)cos(phi) i + sin(theta)sin(phi) j + cos(theta) k, with theta and phi labelled as angles.

Unit vector equation: n-hat = cos(θ)cos(φ) i-hat + cos(θ)sin(φ) j-hat − sin(θ) k-hat, written in spherical polar components.

Unit vector equation: φ̂ = −sin(φ) î + cos(φ) ĵ, written in plain vector notation.

Equation dS_r = r^2 sin θ dθ dφ, the area element on a sphere in spherical polar coordinates, written in plain text.


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].
 
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
  • Handwritten cross product derivation in spherical coordinates: the determinant of unit vectors e1, e2, e3 with R cos θ cos φ, R cos θ sin φ, -R sin θ and related entries, simplifying to R² sin θ e_r.
    Notes.JPG
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