Average Values over Surfaces

Why Average over Surface Area?

A scalar quantity can vary from point to point on a surface. Examples include temperature on a metal sheet, density on a thin coating, and concentration on a curved membrane. To describe the quantity by one representative number, we need an average that gives each equal-sized piece of surface equal weight.

A plain average over parameter values is not always appropriate: equal parameter intervals can correspond to surface pieces of different areas. The correct average weights each value by the area where it occurs. This leads to a simple rule: integrate the scalar quantity over the surface, then divide by the surface’s total area.

Definition: Let \(S\) be a surface of finite, positive area, and let \(f\) be an integrable scalar function on \(S\). The average value of \(f\) over \(S\) is

f_{\mathrm{avg}} = \frac{1}{A(S)}\iint_S f\,dS, \qquad A(S)=\iint_S 1\,dS>0.

Here \(dS\) denotes the surface-area element. The average has the same units as \(f\).

Computing the Average from a Parametrization

Suppose a smooth parametrization \(\mathbf r(u,v)\) describes \(S\) for \((u,v)\) in a parameter domain \(D\). Assume it is regular on the relevant interior and covers the surface once, except possibly along boundary or seam sets of surface area zero. Then the area factor is \(\lVert\mathbf r_u\times\mathbf r_v\rVert\), and the scalar surface integral uses this factor as its weight.

Formula: Under the stated coverage condition,

A(S)=\iint_D \lVert\mathbf r_u\times\mathbf r_v\rVert\,du\,dv, \qquad f_{\mathrm{avg}} = \frac{ \displaystyle\iint_D f(\mathbf r(u,v)) \lVert\mathbf r_u\times\mathbf r_v\rVert\,du\,dv }{ \displaystyle\iint_D \lVert\mathbf r_u\times\mathbf r_v\rVert\,du\,dv }.

If a parametrization covers part of the surface more than once on a set of positive area, these integrals count that part with its multiplicity. Use a single covering or explicitly correct for the multiplicity.

For a graph \(z=g(x,y)\) over a region \(R\), the same rule uses \(dS=\sqrt{1+g_x^2+g_y^2}\,dA\). In either representation, the numerator accumulates the scalar value with area weighting, while the denominator accumulates area alone.

Why the Area Weight Is Necessary

Imagine dividing a surface into many small pieces. A piece of area \(\Delta S_i\) on which the scalar is approximately \(f_i\) contributes approximately \(f_i\Delta S_i\) to the scalar surface integral. The total area is approximately \(\sum_i\Delta S_i\), so the corresponding average is approximately \(\frac{\sum_i f_i\Delta S_i}{\sum_i\Delta S_i}\). Refining the pieces gives the ratio of surface integrals in the definition. A larger surface piece therefore contributes in proportion to its area, not in proportion to how many parameter values happen to describe it.

If the area factor is constant on \(D\), it cancels from the ratio, and the average can be computed as an ordinary average over the parameter domain. If the factor varies, the ordinary parameter average is generally not the surface average. It can still agree for particular functions—for example, for a constant function—but it is not a general replacement for area weighting.

Theorem: If \(f\) is continuous on a compact surface \(S\) of positive area, then its average lies between its minimum and maximum values on \(S\).

To see this, let \(m\) and \(M\) be the minimum and maximum of \(f\). At every point of \(S\), \(m\leq f\leq M\). Integrating preserves these inequalities because \(dS\) is nonnegative:

mA(S) \leq \iint_S f\,dS \leq MA(S).

Dividing by \(A(S)>0\) gives \(m\leq f_{\mathrm{avg}}\leq M\). This is a useful check: an average cannot exceed the range of values being averaged.

Worked Examples

Worked Example: Average Temperature on a Rectangular Plane Patch

A rectangular surface is parametrized by \(\mathbf r(u,v)=(u,v,0)\), where \(0\leq u\leq2\) and \(0\leq v\leq1\). Its temperature is \(T(u,v)=18+2u+3v\) degrees. Find its average temperature.

First compute the tangent vectors and area factor:

\mathbf r_u=(1,0,0),\qquad \mathbf r_v=(0,1,0),\qquad \mathbf r_u\times\mathbf r_v=(0,0,1), \qquad \lVert\mathbf r_u\times\mathbf r_v\rVert=1.

Thus the total area is

A(S)=\int_0^2\int_0^1 1\,dv\,du =\int_0^2 1\,du =2.

The scalar surface integral is

\begin{aligned} \iint_S T\,dS &=\int_0^2\int_0^1(18+2u+3v)\,dv\,du\\ &=\int_0^2\left[18v+2uv+\frac{3}{2}v^2\right]_{0}^{1}\,du\\ &=\int_0^2\left(\frac{39}{2}+2u\right)\,du\\ &=\left[\frac{39}{2}u+u^2\right]_{0}^{2}\\ &=39+4=43. \end{aligned}

Therefore,

T_{\mathrm{avg}}=\frac{43}{2}=21.5\text{ degrees}.

Check: multiplying the average by the area recovers the integral, since \(21.5\cdot2=43\). The units are degrees, as required for an average temperature.

Worked Example: Average Temperature on a Hemisphere

Consider the upper hemisphere of radius \(2\), parametrized by \(\mathbf r(\phi,\theta)=(2\sin\phi\cos\theta,2\sin\phi\sin\theta,2\cos\phi)\), where \(0\leq\phi\leq\pi/2\) and \(0\leq\theta\leq2\pi\). The seam and pole do not affect the area integrals. Let the temperature be \(T=20+3z\) degrees. Find its average.

Here \(z=2\cos\phi\), so \(T=20+6\cos\phi\). The spherical area factor is \(4\sin\phi\). The parametrization covers the hemisphere once apart from its seam and pole, which have zero surface area. The total area is

A(S)=\int_0^{2\pi}\int_0^{\pi/2}4\sin\phi\,d\phi\,d\theta =2\pi\cdot4[-\cos\phi]_0^{\pi/2} =8\pi.

For the numerator,

\begin{aligned} \iint_S T\,dS &=\int_0^{2\pi}\int_0^{\pi/2} (20+6\cos\phi)4\sin\phi\,d\phi\,d\theta\\ &=2\pi\int_0^{\pi/2} (80\sin\phi+24\sin\phi\cos\phi)\,d\phi\\ &=2\pi\left(80+12\right)=184\pi. \end{aligned}

Consequently,

T_{\mathrm{avg}}=\frac{184\pi}{8\pi}=23\text{ degrees}.

Check: \(23\cdot8\pi=184\pi\), the computed temperature integral. Also, \(20\leq T\leq26\) on the upper hemisphere, and \(23\) lies in that range.

Worked Example: Average Density on a Cylindrical Surface

Take the side of a cylinder of radius \(1\), with \(0\leq z\leq3\), parametrized by \(\mathbf r(\theta,z)=(\cos\theta,\sin\theta,z)\), where \(0\leq\theta\leq2\pi\). A coating has surface density \(\rho=2+z\) kilograms per square metre. Find the average density.

The tangent vectors are \(\mathbf r_\theta=(-\sin\theta,\cos\theta,0)\) and \(\mathbf r_z=(0,0,1)\), so \(\lVert\mathbf r_\theta\times\mathbf r_z\rVert=1\). The seam at \(\theta=0,2\pi\) has zero area, and the rest is covered once. Thus

A(S)=\int_0^{2\pi}\int_0^3 1\,dz\,d\theta=6\pi.

The density integral, which is the coating’s total mass, is

\begin{aligned} \iint_S \rho\,dS &=\int_0^{2\pi}\int_0^3(2+z)\,dz\,d\theta\\ &=2\pi\left[2z+\frac{z^2}{2}\right]_0^3\\ &=2\pi\left(6+\frac{9}{2}\right)=21\pi\text{ kilograms}. \end{aligned}

Therefore,

\rho_{\mathrm{avg}}=\frac{21\pi}{6\pi} =\frac{7}{2}=3.5\text{ kilograms per square metre}.

Check: average density times area gives the mass: \(3.5\cdot6\pi=21\pi\) kilograms. The result is also between the minimum density \(2\) and maximum density \(5\).

A Common Pitfall: Averaging Parameter Values Uniformly

On the hemisphere above, the area factor \(4\sin\phi\) varies with \(\phi\). If we incorrectly average the temperature uniformly over the parameter rectangle, we obtain

\frac{1}{\pi^2} \int_0^{2\pi}\int_0^{\pi/2}(20+6\cos\phi)\,d\phi\,d\theta = 20+\frac{12}{\pi},

which is not \(23\). This calculation weights equal intervals of \(\phi\) equally, even though they do not represent equal areas on the hemisphere. Including \(4\sin\phi\) gives the correct surface average. A nonconstant area factor does not force every uniform parameter average to be wrong: for example, if \(f=25\) everywhere, both averages equal \(25\). The point is that surface-area weighting is needed in general.

Another coverage error is to let \(\theta\) run from \(0\) to \(4\pi\) on a full cylinder. That traces its side twice, so the area integral counts every point twice. The ratio for an average may happen to cancel this factor when the entire surface is duplicated uniformly, but the resulting integral is not the actual area or scalar integral. Use a one-to-one covering except on zero-area seams, or account for multiplicity explicitly.

Interpreting the Result

The average is the constant scalar value that would produce the same total scalar integral over a surface of the same area. For temperature, this gives a representative temperature; for density, multiplying the average density by area gives total mass. The average is not generally the value at a distinguished point, nor does it describe how the quantity varies across the surface. Its meaning is an area-weighted summary.

Takeaway: For a finite-area surface with \(A(S)>0\), compute the average scalar value as the scalar surface integral divided by total area. In parameter coordinates, include the surface area factor and ensure the surface is not inadvertently counted multiple times.

Check Your Understanding

Use surface-area weighting in each question, and state the relevant area or average-value check.

A scalar function is constant with value \(c\) on a surface of positive area. What is its surface average, and why?
When can the area factor be cancelled from a parametrized average formula for all scalar functions?
Why does the upper-hemisphere area factor depend on \(\phi\), and what error can result from ignoring it?
A surface has area \(10\) square metres and scalar surface integral \(35\) units. What is the average scalar value, including its units if the integral represents a total quantity?
What must be checked when a parametrization’s parameter domain traces a seam or repeats part of a surface?

What You'll Learn