Surface Integrals of Scalar Functions

Consider a scalar function f (x, y, z) and a surface S. Let S be given by the position vector

r (u, v) = x (u, v) i + y (u, v) j + z (u, v) k,

where the coordinates (u, v) range over some domain D (u, v) of the uv-plane. Notice that the function f (x, y, z) is evaluated only on the points of the surface S, that is

f [r (u, v)] = f [x (u, v), y (u, v), z (u, v)] .

The surface integral of scalar function $f (x, y, z)$ over the surface $S$ is defined as

\iint_{S} f (x, y, z) d S = \iint_{D (u, v)} f (x (u, v), y (u, v), z (u, v)) | \frac{\partial r}{\partial u} \times \frac{\partial r}{\partial v} | d u d v,

where the partial derivatives $\frac{\partial r}{\partial u}$ and $\frac{\partial r}{\partial v}$ are given by

\frac{\partial r}{\partial u} = \frac{\partial x}{\partial u} (u, v) i + \frac{\partial y}{\partial u} (u, v) j + \frac{\partial z}{\partial u} (u, v) k,

\frac{\partial r}{\partial v} = \frac{\partial x}{\partial v} (u, v) i + \frac{\partial y}{\partial v} (u, v) j + \frac{\partial z}{\partial v} (u, v) k,

and $\frac{\partial r}{\partial u} \times \frac{\partial r}{\partial v}$ is the cross product. The vector $\frac{\partial r}{\partial u} \times \frac{\partial r}{\partial v}$ is perpendicular to the surface at the point $r (u, v) .$

The absolute value

d S = | \frac{\partial r}{\partial u} \times \frac{\partial r}{\partial v} | d u d v

is called the area element: it represents the area $d S$ of a small patch of the surface obtained by changing the coordinates $u$ and $v$ by small amounts $d u$ and $d v$ (Figure $1$ ).

The area of the surface $S$ is given by the integral

A = \iint_{S} d S .

If the surface $S$ is defined by the equation $z = z (x, y),$ where $z (x, y)$ is a differentiable function in the domain $D (x, y),$ then the surface integral can be found by the formula

\iint_{S} f (x, y, z) d S = \iint_{D (x, y)} f (x, y, z (x, y)) \sqrt{1 + {(\frac{\partial z}{\partial x})}^{2} + {(\frac{\partial z}{\partial y})}^{2}} d x d y .

If a surface $S$ consists of several "patches" $S_{i},$ then for calculating the surface integral we can apply the additivity property:

\iint_{S} f (x, y, z) d S = \sum_{i = 1}^{n} \iint_{S_{i}} f (x, y, z) d S_{i} .

Solved Problems

Example 1.

Calculate the surface integral $\iint_{S} (x + y + z) d S,$ where $S$ is the portion of the plane $x + 2 y + 4 z = 4$ lying in the first octant $(x \geq 0,$ $y \geq 0,$ $z \geq 0) .$

Solution.

We rewrite the equation of the plane in the form

Find the partial derivatives:

Applying the formula

we can express the surface integral in terms of the double integral:

The region of integration is the triangle shown in Figure

The region of integration is a triangle. — Figure 2.

Calculate the given integral:

Example 2.

Evaluate the surface integral where is the total area of the cone

Solution.

Let be side surface of the cone, and be its base. We can write the given integral as the sum of two integrals:

Find the first integral using the formula

Here the partial derivatives are

Then

Since on the base of the cone, then the domain is defined by the inequality (Figure ).

Surface integral over the cone — Figure 3.

Hence, the integral is written as

In polar coordinates we have

Consider now the second integral The base of the cone is described by the equation Therefore,

where represents the area of the base, which is Then

Thus, the full value of the initial surface integral is

See more problems on Page 2.