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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 \PageIndex{1}

In Figure \text{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 ∫ \textbf{g} \cdot d\textbf{A} through the entire surface of the cylinder. Of course we already know, from Gauss’s theorem, that the answer is = −4 \pi Gm, but we would like to see a surface integral actually carried out.

V19.png
\text{FIGURE V.19}

I have drawn a small element of the surface. Its area δA is dz times aδ \phi, where \phi is the usual azimuthal angle of cylindrical coordinates. That is, δA = a \ δz \ δ\phi. The magnitude g of the field there is Gm/r^2, and the angle between \textbf{g} and d\textbf{A} is 90^\circ + θ. The outward flux through the small element is

\textbf{g} \cdot δ \textbf{A} = \frac{Gma \cos (θ+90^\circ) δz δ \phi}{r^2}.

(This is negative – i.e. it is actually an inward flux – because \cos (θ + 90^\circ ) = −\sin θ.) When integrated around the elemental strip δz, this is - \frac{2 \pi Gma \sin θ δz}{r^2}. 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 = \pi/2 to α and double it, where \tan α = a/h. We’ll need to express z and r in terms of θ (that’s easy:- z = a \cot θ and r = a \csc θ),and the integral becomes

4 \pi Gm \int_{\pi/2}^α \sin θ \ d θ = -4 \pi Gm \cos α \label{5.6.1} \tag{5.6.1}

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

V20.png
\text{FIGURE V.20}

This time, δA = ρ \ δρ \ δ\phi, g = Gm/r^2 and the angle between \textbf{g} and δ\textbf{A} is 180^\circ − θ. The outwards flux through the small element is \frac{Gmρ \cos (180^\circ - θ)δρδ\phi}{r^2} and when integrated around the annulus this becomes -\frac{2\pi Gm \cos θ \ ρδρ}{r^2}. We now have to integrate this from ρ = 0 to a, or, better, from θ = 0 to α. We have r = h \sec θ and ρ = h \tan θ, and the integral becomes

-2 \pi Gm \int_0^α \sin θ d θ = - 2\pi Gm (1- \cos α). \label{5.6.2} \tag{5.6.2}

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

\phi = -4\pi Gm, \label{5.6.3} \tag{5.6.3}

as expected from Gauss’s theorem.

Example \PageIndex{2}

In figure \text{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
\text{FIGURE V.21}

The vector δ\textbf{A} representing the element of area is directed away from the centre of the sphere, and the vector \textbf{g} is directed towards the nearest point of the rod. The angle between them is θ + 90^\circ. The magnitude of δ\textbf{A} in spherical coordinates is a^2 \sin θδθδ \phi, and the magnitude of \textbf{g} is (see Equation 5.4.15) \frac{2Gλ}{a \sin θ}. The dot product \textbf{g}⋅δ \textbf{A} is

\frac{2Gλ}{a \sin θ} \cdot a^2 \sin θδθδ \phi \cdot \cos ( θ + 90^\circ) = - 2 G λ a \sin θδθδ \phi . \label{5.6.4} \tag{5.6.4}

To find the total flux, this must be integrated from \phi = 0 to 2 \pi and from θ = 0 to \pi. The result, as expected, is −8 \pi Gαλ.


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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