Surface integral in Spherical Polar

samjohnny
Messages
83
Reaction score
1

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
    30.8 KB · Views: 406
Physics news on Phys.org
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
    16.6 KB · Views: 401
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top