# Physical Applications of Surface Integrals

Surface integrals are used in multiple areas of physics and engineering. In particular, they are used for calculations of

• mass of a shell;
• center of mass and moments of inertia of a shell;
• gravitational force and pressure force;
• fluid flow and mass flow across a surface;
• electric charge distributed over a surface;
• electric fields (Gauss' Law in electrostatics).

## Mass of a Surface

Let $$S$$ be a smooth thin shell. The mass per unit area of the shell is described by a continuous function $$\mu \left( {x,y,z} \right).$$ Then the total mass of the shell is expressed through the surface integral of scalar function by the formula

$m = \iint\limits_S {\mu \left( {x,y,z} \right)dS} .$

## Center of Mass and Moments of Inertia of a Surface

Let a mass $$m$$ be distributed over a thin shell $$S$$ with a continuous density function $$\mu \left( {x,y,z} \right).$$ The coordinates of the center of mass of the shell are defined by the formulas

${x_C} = \frac{{{M_{yz}}}}{m},\;\; {y_C} = \frac{{{M_{xz}}}}{m},\;\; {z_C} = \frac{{{M_{xy}}}}{m},$

where

${M_{yz}} = \iint\limits_S {x\mu \left( {x,y,z} \right)dS} ,\;\; {M_{xz}} = \iint\limits_S {y\mu \left( {x,y,z} \right)dS} ,\;\; {M_{xy}} = \iint\limits_S {z\mu \left( {x,y,z} \right)dS}$

are so-called the first moments about the coordinate planes $$x = 0,$$ $$y = 0,$$ and $$z = 0,$$ respectively.

The moments of inertia about the $$x-,$$ $$y-,$$ and $$z-$$axis are given by

${I_x} = \iint\limits_S {\left( {{y^2} + {z^2}} \right)\mu \left( {x,y,z} \right)dS} ,\;\; {I_y} = \iint\limits_S {\left( {{x^2} + {z^2}} \right)\mu \left( {x,y,z} \right)dS} ,\;\; {I_z} = \iint\limits_S {\left( {{x^2} + {y^2}} \right)\mu \left( {x,y,z} \right)dS}.$

The moments of inertia of a shell about the $$xy-,$$ $$yz-,$$ and $$xz-$$plane are defined by the formulas

${I_{xy}} = \iint\limits_S {{z^2}\mu \left( {x,y,z} \right)dS} ,\;\; {I_{yz}} = \iint\limits_S {{x^2}\mu \left( {x,y,z} \right)dS} ,\;\; {I_{xz}} = \iint\limits_S {{y^2}\mu \left( {x,y,z} \right)dS} .$

## Gravitational Force

Let $$m$$ be a mass at a point $$\left( {{x_0},{y_0},{z_0}} \right)$$ outside the surface $$S$$ (Figure $$1$$).

Then the force of attraction between the surface $$S$$ and the mass $$m$$ is given by

$\mathbf{F} = Gm\iint\limits_S {\mu \left( {x,y,z} \right)\frac{\mathbf{r}}{{{r^3}}}dS} ,$

where $$\mathbf{r} =$$ $$\left( {x - {x_0},y - {y_0},z - {z_0}} \right),$$ $$G$$ is gravitational constant, $${\mu \left( {x,y,z} \right)}$$ is the density function.

## Pressure Force

Suppose a surface $$S$$ be given by the position vector $$\mathbf{r}$$ and is stressed by a pressure force acting on it. Examples of such surfaces are dams, aircraft wings, compressed gas storage tanks, etc. The total force $$\mathbf{F}$$ created by the pressure $$p\left( \mathbf{r} \right)$$ is given by the surface integral

$\mathbf{F} = \iint\limits_S {p\left( \mathbf{r} \right)d\mathbf{S}} .$

By definition, the pressure is directed in the direction of the normal of $$S$$ in each point. Therefore, we can write:

$\mathbf{F} = \iint\limits_S {p\left( \mathbf{r} \right)d\mathbf{S}} = \iint\limits_S {p\mathbf{n}dS} ,$

where $$\mathbf{n}$$ is the unit normal vector to the surface $$S.$$

## Fluid Flux and Mass Flux

If the vector field is the fluid velocity $$\mathbf{v}\left( \mathbf{r} \right),$$ the flux across a surface $$S$$ is called the fluid flux. It is equal to the volume of the fluid passing across $$S$$ per unit time and is given by

$\Phi = \iint\limits_S {\mathbf{v}\left( \mathbf{r} \right) \cdot d\mathbf{S}} .$

Similarly, the flux of the vector field $$\mathbf{F} = \rho \mathbf{v},$$ where $$\rho$$ is the fluid density, is called the mass flux and is given by

$\Phi = \iint\limits_S {\rho \mathbf{v}\left( \mathbf{r} \right) \cdot d\mathbf{S}} .$

It is equal to the mass passing across a surface $$S$$ per unit time.

## Surface Charge

Let $$\sigma \left( {x,y} \right)$$ be the surface charge density. The total amount of charge distributed over the conducting surface $$S$$ is expressed by the formula

$Q = \iint\limits_S {\sigma \left( {x,y} \right)dS} .$

## Gauss' Law

The electric flux $$\mathbf{D}$$ through any closed surface $$S$$ is proportional to the charge $$Q$$ enclosed by the surface:

$\Phi = \iint\limits_S {\mathbf{D} \cdot d\mathbf{S}} = \sum\limits_i {{Q_i}} ,$

where $$\mathbf{D} = \varepsilon {\varepsilon _0}\mathbf{E},$$ $$\mathbf{E}$$ is the magnitude of the electric field strength, $$\varepsilon$$ is permittivity of material, and $${\varepsilon _0} = 8,85\; \times$$ $${10^{ - 12}}\,\text{F/m}$$ is permittivity of free space.

For the discrete case the total charge $$Q$$ is the sum over all the enclosed charges.

Gauss' Law is a general law applying to any closed surface. For geometries of sufficient symmetry, it simplifies the calculation of electric field. Gauss' Law is the first of Maxwell's equations, the four fundamental equations for electricity and magnetism.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the mass of cylindrical surface parameterized by

$\mathbf{r}\left( {u,v} \right) = a\cos u \cdot \mathbf{i} + a\sin u \cdot \mathbf{j} + v \cdot \mathbf{k},$

where $$0 \le u \le 2\pi ,$$ $$0 \le v \le H$$ (Figure $$2$$). The surface density is defined by the function $\mu \left( {x,y,z} \right) = {z^2}\left( {{x^2} + {y^2}} \right).$

### Example 2

Find the mass of the parabolic surface $z = {x^2} + {y^2}, 0 \le z \le 1$ with density $\mu \left( {x,y,z} \right) = z.$

### Example 1.

Find the mass of cylindrical surface parameterized by

$\mathbf{r}\left( {u,v} \right) = a\cos u \cdot \mathbf{i} + a\sin u \cdot \mathbf{j} + v \cdot \mathbf{k},$

where $$0 \le u \le 2\pi ,$$ $$0 \le v \le H$$ (Figure $$2$$). The surface density is defined by the function $\mu \left( {x,y,z} \right) = {z^2}\left( {{x^2} + {y^2}} \right).$

Solution.

The mass of the surface is given by the formula

$m = \iint\limits_S {\mu \left( {x,y,z} \right)dS} .$

Calculate the area element $$dS:$$

$dS = \left| {\frac{{\partial \mathbf{r}}}{{\partial u}} \times \frac{{\partial \mathbf{r}}}{{\partial v}}} \right|dudv.$

Find the partial derivatives and their cross product:

$\frac{{\partial \mathbf{r}}}{{\partial u}} = - a\sin u \cdot \mathbf{i} + a\cos u \cdot \mathbf{j} + 0 \cdot \mathbf{k},$
$\frac{{\partial \mathbf{r}}}{{\partial v}} = 0 \cdot \mathbf{i} + 0 \cdot \mathbf{j} + 1 \cdot \mathbf{k},$
$\Rightarrow \frac{{\partial \mathbf{r}}}{{\partial u}} \times \frac{{\partial \mathbf{r}}}{{\partial v}} = \left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ { - a\sin u} & {a\cos u} & 0\\ 0 & 0 & 1 \end{array}} \right| = a\cos u \cdot \mathbf{i} + a\sin u \cdot \mathbf{j},$
$\Rightarrow \left| {\frac{{\partial \mathbf{r}}}{{\partial u}} \times \frac{{\partial \mathbf{r}}}{{\partial v}}} \right| = \sqrt {{a^2}{{\cos }^2}u + {a^2}{{\sin }^2}u} = a.$

So that $$dS = adudv.$$ Then the mass of the surface is

$m = \iint\limits_S {\mu \left( {x,y,z} \right)dS} = \iint\limits_S {{z^2}\left( {{x^2} + {y^2}} \right)dS} = \iint\limits_{D\left( {u,v} \right)} {{v^2}\left( {{a^2}{{\cos }^2}u + {a^2}{{\sin }^2}u} \right)adudv} = {a^3}\int\limits_0^{2\pi } {du} \int\limits_0^H {{v^2}dv} = 2\pi {a^3}\int\limits_0^H {{v^2}dv} = 2\pi {a^3}\left[ {\left. {\left( {\frac{{{v^3}}}{3}} \right)} \right|_0^H} \right] = \frac{{2\pi {a^3}{H^3}}}{3}.$

### Example 2.

Find the mass of the parabolic surface $z = {x^2} + {y^2}, 0 \le z \le 1$ with density $\mu \left( {x,y,z} \right) = z.$

Solution.

We use the formula

$m = \iint\limits_S {\mu \left( {x,y,z} \right)dS}.$

The projection $$D\left( {x,y} \right)$$ of the parabolic surface $$S$$ onto the $$xy$$-plane is the circle of radius $$1$$ centered at the origin. Hence, we can write:

$m = \iint\limits_S {\mu \left( {x,y,z} \right)dS} = \iint\limits_S {zdS} = \iint\limits_{D\left( {x,y} \right)} {\left( {{x^2} + {y^2}} \right) \sqrt {1 + 4{x^2} + 4{y^2}} dxdy}.$

By changing to polar coordinates, we obtain

$m = \iint\limits_{D\left( {r,\varphi } \right)} {{r^2}\sqrt {1 + 4{r^2}} rdrd\varphi } = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^1 {{r^3}\sqrt {1 + 4{r^2}} dr} = 2\pi \int\limits_0^1 {{r^3}\sqrt {1 + 4{r^2}} dr} .$

Make the substitution $$1 + 4{r^2} = {u^2}.$$ Then $$8rdr = 2udu$$ or $$rdr = {\frac{{udu}}{4}}.$$ Here $$u = 1$$ when $$r = 0,$$ and $$u = \sqrt 5$$ when $$r = 1.$$ Hence, the integral becomes

$m = 2\pi \int\limits_1^{\sqrt 5 } {\frac{{{u^2} - 1}}{4}\sqrt {{u^2}} \frac{{udu}}{4}} = \frac{\pi }{8}\int\limits_1^{\sqrt 5 } {\left( {{u^2} - 1} \right){u^2}du} = \frac{\pi }{8}\int\limits_1^{\sqrt 5 } {\left( {{u^4} - {u^2}} \right)du} = \frac{\pi }{8}\left[ {\left. {\left( {\frac{{{u^5}}}{5} - \frac{{{u^3}}}{3}} \right)} \right|_1^{\sqrt 5 }} \right] = \frac{\pi }{8}\left[ {\left( {\frac{{{{\left( {\sqrt 5 } \right)}^5}}}{5} - \frac{{{{\left( {\sqrt 5 } \right)}^3}}}{3}} \right) - \left( {\frac{1}{5} - \frac{1}{3}} \right)} \right] = \frac{\pi }{8}\left( {\frac{{10\sqrt 5 }}{3} + \frac{2}{{15}}} \right) = \frac{{\pi \left( {25\sqrt 5 + 1} \right)}}{{60}}.$

See more problems on Page 2.