5.6: Calculating Surface Integrals

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.

$\text{FIGURE V.19}$

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.

$\text{FIGURE V.20}$

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 $−8πGαλ$.

$\text{FIGURE V.21}$

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 $−8 \pi Gαλ$.