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 $μ (x, y, z) .$ Then the total mass of the shell is expressed through the surface integral of scalar function by the formula

m = \iint_{S} μ (x, y, z) d S .

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 $μ (x, y, z) .$ The coordinates of the center of mass of the shell are defined by the formulas

x_{C} = \frac{M_{y z}}{m}, y_{C} = \frac{M_{x z}}{m}, z_{C} = \frac{M_{x y}}{m},

where

M_{y z} = \iint_{S} x μ (x, y, z) d S, M_{x z} = \iint_{S} y μ (x, y, z) d S, M_{x y} = \iint_{S} z μ (x, y, z) d S

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_{S} (y^{2} + z^{2}) μ (x, y, z) d S, I_{y} = \iint_{S} (x^{2} + z^{2}) μ (x, y, z) d S, I_{z} = \iint_{S} (x^{2} + y^{2}) μ (x, y, z) d S .

The moments of inertia of a shell about the $x y -,$ $y z -,$ and $x z -$ plane are defined by the formulas

I_{x y} = \iint_{S} z^{2} μ (x, y, z) d S, I_{y z} = \iint_{S} x^{2} μ (x, y, z) d S, I_{x z} = \iint_{S} y^{2} μ (x, y, z) d S .

Gravitational Force

Let $m$ be a mass at a point $(x_{0}, y_{0}, z_{0})$ outside the surface $S$ (Figure $1$ ).

The force of attraction between the surface S and the mass m — Figure 1.

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

where is gravitational constant, is the density function.

Pressure Force

Suppose a surface be given by the position vector 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 created by the pressure is given by the surface integral

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

where is the unit normal vector to the surface

Fluid Flux and Mass Flux

If the vector field is the fluid velocity the flux across a surface is called the fluid flux. It is equal to the volume of the fluid passing across per unit time and is given by

Similarly, the flux of the vector field where is the fluid density, is called the mass flux and is given by

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

Surface Charge

Let be the surface charge density. The total amount of charge distributed over the conducting surface is expressed by the formula

Gauss' Law

The electric flux through any closed surface is proportional to the charge enclosed by the surface:

where is the magnitude of the electric field strength, is permittivity of material, and is permittivity of free space.

For the discrete case the total charge 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

Example 1.

Find the mass of cylindrical surface parameterized by

where (Figure ). The surface density is defined by the function

Solution.

The mass of the surface is given by the formula

Calculate the area element

Find the partial derivatives and their cross product:

So that Then the mass of the surface is

Example 2.

Find the mass of the parabolic surface with density

Solution.

We use the formula

The projection of the parabolic surface onto the -plane is the circle of radius centered at the origin. Hence, we can write:

By changing to polar coordinates, we obtain

Make the substitution Then or Here when and when Hence, the integral becomes

See more problems on Page 2.