Surface Mass from a Density Function

From Surface Density to Mass

A thin curved lamina can have different amounts of mass in different regions. Its mass depends both on the area of each region and on how much mass is packed into each unit of that area. Surface integration brings those two ingredients together.

The key distinction is between area and mass. An area element measures a small piece of the surface; multiplying that area by the local surface density gives the approximate mass of that piece. Adding these contributions over the whole lamina leads to a scalar surface integral.

Definition: A surface density $\rho$ assigns mass per unit area to points of a surface $S$. For a physical lamina, $\rho$ is nonnegative and is measured, for example, in kilograms per square metre. If $\rho$ is integrable on a piecewise smooth surface, the lamina's mass is $$ m=\iint_S \rho\,dS. $$

Here $dS$ denotes surface area, not mass. The density must be included in the integral for the result to represent mass. If density is constant, the formula reduces to that density multiplied by the surface area.

Why the Mass Formula Works

Partition the lamina into many small surface pieces. On a sufficiently small piece, the density is approximately constant, so its mass is approximately the density at a representative point times the piece's area. Summing those estimates and refining the partition gives the integral.

Approximate each small piece.
If the $i$-th piece has area $\Delta S_i$ and representative point $\mathbf{p}_i$, its mass is approximately $\rho(\mathbf{p}_i)\Delta S_i$.

Add the contributions.
The total mass is approximated by $\sum_i \rho(\mathbf{p}_i)\Delta S_i$.

Pass to the limit.
For an integrable density on a suitable piecewise smooth surface, refining the partition gives the surface integral $\iint_S\rho\,dS$.

This reasoning also explains why the area factor is essential. A parameter rectangle generally does not have the same area as its curved image on the surface. The surface area element corrects for that difference.

Mass on a Parametrized Surface

Suppose a surface patch is parametrized by $\mathbf{r}(u,v)$, for $(u,v)$ in a parameter domain $D$. Assume the parametrization is continuously differentiable, regular on the interior of the domain, and covers the patch without counting its interior more than once. Its area element is

dS=\left\lVert\mathbf{r}_u\times\mathbf{r}_v\right\rVert\,du\,dv.

If the density is given as a function $\rho(x,y,z)$ of position, then at parameter values $(u,v)$ the density is $\rho(\mathbf{r}(u,v))$. Therefore, the mass is

m=\iint_D \rho(\mathbf{r}(u,v)) \left\lVert\mathbf{r}_u\times\mathbf{r}_v\right\rVert\,du\,dv.

The cross-product magnitude converts a small parameter rectangle into its approximate surface area. The density is evaluated at the corresponding point on the surface. Reversing the order of the two parameters changes the direction of the cross product but not its magnitude, so it does not change this mass integral.

Mass on a Graph

For a graph $z=f(x,y)$ over a region $R$ in the $xy$-plane, use $\mathbf{r}(x,y)=(x,y,f(x,y))$. The area factor is $\sqrt{1+f_x^2+f_y^2}$. Thus, if the density is specified in three-dimensional coordinates, the mass is

m=\iint_R \rho(x,y,f(x,y)) \sqrt{1+f_x^2+f_y^2}\,dx\,dy.

This formula requires a differentiable graph over the region and an integrable density. The density must be evaluated at the graph height, and the square-root factor must be retained unless it happens to equal $1$. Since $dx\,dy$ measures projected area in the $xy$-plane, it is not generally the actual surface area.

Worked Example: Constant Density on a Sloping Graph

Find the mass of the graph $z=x+y$ over $0\leq x\leq1\,\mathrm{m}$, $0\leq y\leq1\,\mathrm{m}$, when the density is constant at $2\,\mathrm{kg/m^2}$.

First compute the partial derivatives: $f_x=1$ and $f_y=1$. Therefore the graph area factor is

\sqrt{1+f_x^2+f_y^2} =\sqrt{1+1^2+1^2} =\sqrt{3}.

Include the density in the mass integral. With $x$ and $y$ measured in metres, the calculation is

\begin{aligned} m &=\int_0^{1\,\mathrm{m}}\int_0^{1\,\mathrm{m}} (2\,\mathrm{kg/m^2})\sqrt{3}\,dy\,dx\\ &=(2\,\mathrm{kg/m^2})\sqrt{3} (1\,\mathrm{m})(1\,\mathrm{m})\\ &=2\sqrt{3}\,\mathrm{kg}. \end{aligned}

As a check, the graph's area is $\sqrt{3}(1\,\mathrm{m})(1\,\mathrm{m})=\sqrt{3}\,\mathrm{m^2}$. Multiplying by $2\,\mathrm{kg/m^2}$ gives $2\sqrt{3}\,\mathrm{kg}$, as above. The units are mass, not area.

Nonuniform Density on a Parametrized Patch

For a nonuniform lamina, the density varies from point to point. When it is given as a function of position, substitute the parametrization into that function before integrating. The area factor remains present: density variation does not replace the geometric correction.

Worked Example: A Cylinder with Density Increasing in Height

Consider the quarter of a circular cylinder of radius $2\,\mathrm{m}$ with $0\leq\theta\leq\pi/2$ and $0\leq z\leq3\,\mathrm{m}$, parametrized by

\mathbf{r}(\theta,z) =(2\,\mathrm{m}\cos\theta,\;2\,\mathrm{m}\sin\theta,\;z).

Let its density be $\rho(x,y,z)=2\,\mathrm{kg/m^2}\left(1+\frac{z}{3\,\mathrm{m}}\right)$. Compute the mass.

The tangent vectors and their cross product are

\begin{aligned} \mathbf{r}_\theta &=(-2\,\mathrm{m}\sin\theta,\;2\,\mathrm{m}\cos\theta,\;0),\\ \mathbf{r}_z&=(0,0,1),\\ \mathbf{r}_\theta\times\mathbf{r}_z &=(2\,\mathrm{m}\cos\theta,\;2\,\mathrm{m}\sin\theta,\;0). \end{aligned}

Consequently, the area factor is

\left\lVert\mathbf{r}_\theta\times\mathbf{r}_z\right\rVert =2\,\mathrm{m}\sqrt{\cos^2\theta+\sin^2\theta} =2\,\mathrm{m}.

Substituting the density and the area factor gives

\begin{aligned} m &=\int_0^{\pi/2}\int_0^{3\,\mathrm{m}} 2\,\frac{\mathrm{kg}}{\mathrm{m^2}} \left(1+\frac{z}{3\,\mathrm{m}}\right) (2\,\mathrm{m})\,dz\,d\theta\\ &=4\,\frac{\mathrm{kg}}{\mathrm{m}} \int_0^{\pi/2}\int_0^{3\,\mathrm{m}} \left(1+\frac{z}{3\,\mathrm{m}}\right)\,dz\,d\theta. \end{aligned}

Evaluate the inner integral, keeping the endpoint values explicit:

\begin{aligned} \int_0^{3\,\mathrm{m}} \left(1+\frac{z}{3\,\mathrm{m}}\right)\,dz &=\left[z+\frac{z^2}{6\,\mathrm{m}}\right]_0^{3\,\mathrm{m}}\\ &=3\,\mathrm{m}+\frac{9\,\mathrm{m^2}}{6\,\mathrm{m}}\\ &=3\,\mathrm{m}+\frac{3}{2}\,\mathrm{m} =\frac{9}{2}\,\mathrm{m}. \end{aligned}

The remaining integral is

m=4\,\frac{\mathrm{kg}}{\mathrm{m}} \left(\frac{9}{2}\,\mathrm{m}\right) \left(\frac{\pi}{2}\right) =9\pi\,\mathrm{kg}.

To check, split the density into its constant part and its height-dependent part. The constant part contributes $2\,\mathrm{kg/m^2}$ times the cylinder patch area $2\,\mathrm{m}\cdot(3\,\mathrm{m})\cdot(\pi/2)=3\pi\,\mathrm{m^2}$, or $6\pi\,\mathrm{kg}$. The added density increases linearly from zero to $2\,\mathrm{kg/m^2}$, so its integral over height contributes half the constant-density amount: $3\pi\,\mathrm{kg}$. Together these give $9\pi\,\mathrm{kg}$, agreeing with the direct calculation.

A Nonuniform Density on a Graph

A useful check on a graph calculation is to track both the density's units and the graph area factor. The next example uses a radial density, meaning its value depends on the distance from the origin in the projected region.

Worked Example: Radial Density on a Paraboloid Patch

Find the mass of the graph $z=(x^2+y^2)/(2\,\mathrm{m})$ above the disk $x^2+y^2\leq(1\,\mathrm{m})^2$, if

\rho(x,y,z)= \left(1+\frac{x^2+y^2}{(1\,\mathrm{m})^2}\right)\mathrm{kg/m^2}.

Write $r=\sqrt{x^2+y^2}$. The graph derivatives are $f_x=x/(1\,\mathrm{m})$ and $f_y=y/(1\,\mathrm{m})$, so the area factor is $\sqrt{1+r^2/(1\,\mathrm{m})^2}$. In polar coordinates, the projected area element is $r\,dr\,d\theta$, and the density on the graph is $(1+r^2/(1\,\mathrm{m})^2)\,\mathrm{kg/m^2}$. Hence

m=\int_0^{2\pi}\int_0^{1\,\mathrm{m}} \left(1+\frac{r^2}{(1\,\mathrm{m})^2}\right)^{3/2} \frac{\mathrm{kg}}{\mathrm{m^2}}\,r\,dr\,d\theta.

Set $t=r/(1\,\mathrm{m})$. Then $r\,dr=(1\,\mathrm{m})^2t\,dt$, and the limits for $t$ are $0$ and $1$. Therefore

\begin{aligned} m &=2\pi\,\mathrm{kg}\int_0^1 t(1+t^2)^{3/2}\,dt\\ &=2\pi\,\mathrm{kg}\cdot\frac{1}{2} \int_1^2 u^{3/2}\,du \qquad (u=1+t^2,\;du=2t\,dt)\\ &=2\pi\,\mathrm{kg}\cdot\frac{1}{5} \left[u^{5/2}\right]_1^2\\ &=\frac{2\pi}{5}(2^{5/2}-1)\,\mathrm{kg}. \end{aligned}

For a check of the antiderivative, $\frac{d}{du}\left(\frac{2}{5}u^{5/2}\right)=u^{3/2}$, so the factor $\frac12$ from $t\,dt=du/2$ produces the required $\frac15u^{5/2}$. The unit check also works: the density contributes $\mathrm{kg/m^2}$, and $r\,dr$ contributes $\mathrm{m^2}$, leaving kilograms.

Common Pitfall: Confusing Area with Mass

A frequent error is to calculate the surface area correctly and then report it as the mass. For the sloping graph in the first example, the expression

\int_0^{1\,\mathrm{m}}\int_0^{1\,\mathrm{m}} \sqrt{3}\,dy\,dx=\sqrt{3}\,\mathrm{m^2}

is an area calculation. It does not include the stated density $2\,\mathrm{kg/m^2}$, so it cannot equal the mass. The corrected mass integral is

\int_0^{1\,\mathrm{m}}\int_0^{1\,\mathrm{m}} (2\,\mathrm{kg/m^2})\sqrt{3}\,dy\,dx =2\sqrt{3}\,\mathrm{kg}.

Dimensional analysis is a reliable way to catch this omission: the density times $dS$ must have units of mass. A result in square metres is an area, not a mass.

Setting Up a Surface-Mass Integral

Before evaluating an integral, identify what each factor represents. A dependable setup follows this sequence:

Specify the surface and the region that represents the lamina.
Choose a parametrization or express the surface as a graph.
Write the density at points on that surface.
Compute the correct surface area factor.
Form the integral of density times area element over the parameter domain or projected region.
Evaluate the integral and check that the resulting units are units of mass.

The method applies to nonnegative integrable densities on regular surface patches. If a surface is covered by more than one parametrized patch, take care not to count overlapping area more than once; if it is divided into nonoverlapping patches, add their masses. The central principle remains unchanged: each surface element contributes its local density multiplied by its actual area.

Check Your Understanding

For each question, state what the integrand represents before carrying out any calculation.

A parametrized lamina has density $\rho(\mathbf{r}(u,v))$. What area factor converts $du\,dv$ into surface area?
Why must a density given in three-dimensional coordinates be evaluated at the graph height in a graph-surface mass integral?
A surface area calculation gives $5\,\mathrm{m^2}$, and the lamina has constant density $3\,\mathrm{kg/m^2}$. What is its mass?
For a graph over a planar region, when can $dx\,dy$ itself equal the surface area element?
A proposed mass integral has units of square metres. What likely ingredient is missing or misinterpreted?