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Physics LibreTexts

5.6: Calculating Surface Integrals

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While the concept of a surface integral sounds easy enough, how do we actually calculate one in practice? In this section I do two examples.

Example 5.6.1

In Figure V.19 I show a small mass m, and I have surrounded it with a cylinder of radius a and height 2h. The problem is to calculate the surface integral gdA through the entire surface of the cylinder. Of course we already know, from Gauss’s theorem, that the answer is =4πGm, but we would like to see a surface integral actually carried out.

V19.png
FIGURE V.19

I have drawn a small element of the surface. Its area δA is dz times aδϕ, where ϕ is the usual azimuthal angle of cylindrical coordinates. That is, δA=a δz δϕ. The magnitude g of the field there is Gm/r2, and the angle between g and dA is 90+θ. The outward flux through the small element is

gδA=Gmacos(θ+90)δzδϕr2.

(This is negative – i.e. it is actually an inward flux – because cos(θ+90)=sinθ.) When integrated around the elemental strip δz, this is 2πGmasinθδzr2. To find the flux over the total curved surface, let’s integrate this from z=0 to h and double it, or, easier, from θ=π/2 to α and double it, where tanα=a/h. We’ll need to express z and r in terms of θ (that’s easy:- z=acotθ and r=acscθ),and the integral becomes

4πGmαπ/2sinθ dθ=4πGmcosα

Let us now find the flux through one of the flat ends of the cylinder.

V20.png
FIGURE V.20

This time, δA=ρ δρ δϕ, g=Gm/r2 and the angle between g and δA is 180θ. The outwards flux through the small element is Gmρcos(180θ)δρδϕr2 and when integrated around the annulus this becomes 2πGmcosθ ρδρr2. We now have to integrate this from ρ=0 to a, or, better, from θ=0 to α. We have r=hsecθ and ρ=htanθ, and the integral becomes

2πGmα0sinθdθ=2πGm(1cosα).

There are two ends, so the total flux through the entire cylinder is twice this plus Equation 5.6.1 to give

ϕ=4πGm,

as expected from Gauss’s theorem.

Example 5.6.2

In figure V.21 I have drawn (part of) an infinite rod whose mass per unit length is λ. I have drawn around it a sphere of radius a. The problem will be to determine the total normal flux through the sphere. From Gauss’s theorem, we know that the answer must be 8πGαλ.

V21.png
FIGURE V.21

The vector δA representing the element of area is directed away from the centre of the sphere, and the vector g is directed towards the nearest point of the rod. The angle between them is θ+90. The magnitude of δA in spherical coordinates is a2sinθδθδϕ, and the magnitude of g is (see Equation 5.4.15) 2Gλasinθ. The dot product gδA is

2Gλasinθa2sinθδθδϕcos(θ+90)=2Gλasinθδθδϕ.

To find the total flux, this must be integrated from ϕ=0 to 2π and from θ=0 to π. The result, as expected, is 8πGαλ.


This page titled 5.6: Calculating Surface Integrals is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Jeremy Tatum via source content that was edited to the style and standards of the LibreTexts platform.

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