Quick question - Flux through a sphere

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Homework Help Overview

The problem involves evaluating the surface integral of the vector field r over a sphere of radius a, specifically ∫∫r·n dS, where n is the normal unit vector to the surface. The context is within vector calculus and surface integrals.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of polar coordinates for the surface integral and the correct expression for the surface element dS. There is uncertainty about the limits of integration and the evaluation of the integral.

Discussion Status

Some participants have provided corrections regarding the expression for dS and the limits of integration. There is ongoing exploration of the implications of these corrections on the evaluation of the integral, but no consensus has been reached on the final answer.

Contextual Notes

Participants note potential confusion regarding the angle conventions used in spherical coordinates, which may affect the setup of the integral. There is also mention of the relationship between the dot product r·n and the surface area of the sphere.

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Homework Statement



Evaluate ∫∫r.ndS where r=(x,y,z) and n is a normal unit vector to the surface S, which is a sphere of radius a centred on the origin.

2. The attempt at a solution

I decided to use polar coordinates. The radius of the sphere is clearly constant, a. So a surface element is dS = a2dθdø.

r = a(sinθcosø, sinθsinø, cosθ)

n = (sinθcosø, sinθsinø, cosθ)

r.n = a

therefore, ∫∫r.ndS = ∫∫a3dθdø

where θ varies from 0 to 2π, and ø varies from 0 to π.

This gives an answer of 2π2a3. Is this correct? I'm not so sure.
 
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You forgot that
\mathrm{d} S=a^2 \sin \theta \mathrm{d} \theta \mathrm{d} \phi.
 
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Good point, thank you.

But then I get ∫∫a3sinθdθdø

Which evaluates to zero. Are the limits incorrect?
 
Zatman said:
Good point, thank you.

But then I get ∫∫a3sinθdθdø

Which evaluates to zero. Are the limits incorrect?

Yes: If z = r\cos\theta (there are two exactly opposite conventions for the angle coordinates) then dS = r^2 \sin\theta\,d\theta\,d\phi and the sphere is given by 0 \leq \theta \leq \pi and 0 \leq \phi \leq 2\pi.

Also: once you have that \mathbf{r} \cdot \mathbf{n} = a you know that the answer must be a times the surface area of a sphere of radius a.
 
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Oh, I had theta and phi mixed up. So that gives

∫∫r.ndS = ∫(2a3)dø

from 0 to 2π, hence

= 4πa3
 

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