Surface Integrals of Scalar Functions

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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 $u v$ -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

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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 $(x \geq 0,$ $y \geq 0,$ $z \geq 0) .$

Example 2

Evaluate the surface integral $\iint_{S} z^{2} d S,$ where $S$ is the total area of the cone $\sqrt{x^{2} + y^{2}} \leq z \leq 2.$

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

z = z (x, y) = 1 - \frac{x}{4} - \frac{y}{2} .

Find the partial derivatives:

\frac{\partial z}{\partial x} = - \frac{1}{4}, \frac{\partial z}{\partial y} = - \frac{1}{2} .

Applying the formula

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

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

I = \iint_{S} (x + y + z) d S = \iint_{D} (x + y + 1 - \frac{x}{4} - \frac{y}{2}) \sqrt{1 + {(- \frac{1}{4})}^{2} + {(- \frac{1}{2})}^{2}} d x d y = \iint_{D} (\frac{3 x}{4} + \frac{y}{2} + 1) \frac{\sqrt{21}}{4} d x d y .

The region of integration $D$ is the triangle shown in Figure $2.$

Calculate the given integral:

I = \frac{\sqrt{21}}{4} \int_{0}^{2} [\int_{0}^{4 - 2 y} (\frac{3 x}{4} + \frac{y}{2} + 1) d x] d y = \frac{\sqrt{21}}{16} \int_{0}^{2} [\int_{0}^{4 - 2 y} (3 x + 2 y + 4) d x] d y = \frac{\sqrt{21}}{16} \int_{0}^{2} [(\frac{3 x^{2}}{2} + 2 y x + {4 x) |}_{x = 0}^{4 - 2 y}] d y = \frac{\sqrt{21}}{16} \int_{0}^{2} [\frac{3}{2} {(4 - 2 y)}^{2} + 2 (4 - 2 y) y + 4 (4 - 2 y)] d y = \frac{\sqrt{21}}{32} \int_{0}^{2} [3 {(16 - 16 y + 4 y^{2})}^{2} + 16 y - 8 y^{2} + 32 - 16 y] d y = \frac{\sqrt{21}}{32} \int_{0}^{2} (80 - 48 y + 4 y^{2}) d y = \frac{\sqrt{21}}{4} \int_{0}^{2} (20 - 12 y + y^{2}) d y = \frac{\sqrt{21}}{4} [{(20 y - 6 y^{2} + \frac{y^{3}}{3}) |}_{0}^{2}] = \frac{\sqrt{21}}{4} (40 - 24 + \frac{8}{3}) = \frac{7 \sqrt{21}}{3} .

Example 2.

Evaluate the surface integral $\iint_{S} z^{2} d S,$ where $S$ is the total area of the cone $\sqrt{x^{2} + y^{2}} \leq z \leq 2.$

Solution.

Let $S_{1}$ be side surface of the cone, and $S_{2}$ be its base. We can write the given integral as the sum of two integrals:

I = I_{1} + I_{2} = \iint_{S_{1}} z^{2} d S_{1} + \iint_{S_{2}} z^{2} d S_{2} .

Find the first integral $I_{1},$ using the formula

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

Here the partial derivatives are

\frac{\partial z}{\partial x} = \frac{\partial}{\partial x} \sqrt{x^{2} + y^{2}} = \frac{x}{\sqrt{x^{2} + y^{2}}},

\frac{\partial z}{\partial y} = \frac{\partial}{\partial y} \sqrt{x^{2} + y^{2}} = \frac{y}{\sqrt{x^{2} + y^{2}}} .

Then

\sqrt{1 + {(\frac{\partial z}{\partial x})}^{2} + {(\frac{\partial z}{\partial y})}^{2}} = \sqrt{1 + \frac{x^{2}}{x^{2} + y^{2}} + \frac{y^{2}}{x^{2} + y^{2}}} = \sqrt{2} .

Since $z = 2$ on the base of the cone, then the domain $D (x, y)$ is defined by the inequality $z^{2} + y^{2} \leq 4$ (Figure $3$ ).

Hence, the integral $I_{1}$ is written as

I_{1} = \iint_{S_{1}} z^{2} d S_{1} = \sqrt{2} \iint_{D (x, y)} (x^{2} + y^{2}) d x d y .

In polar coordinates we have

I_{1} = \sqrt{2} \iint_{D (x, y)} (x^{2} + y^{2}) d x d y = \sqrt{2} \int_{0}^{2 π} \int_{0}^{2} r^{2} \cdot r d r d φ = \sqrt{2} \int_{0}^{2 π} d φ \int_{0}^{2} r^{3} d r = \sqrt{2} \cdot 2 π \cdot [{(\frac{r^{4}}{4}) |}_{0}^{2}] = \frac{\sqrt{2} π}{2} (2^{4} - 0) = 8 \sqrt{2} π .

Consider now the second integral $I_{2} .$ The base of the cone is described by the equation $z = 2.$ Therefore,

I_{2} = \iint_{S_{2}} 2^{2} d S_{2} = 4 \iint_{S_{2}} d S_{2},

where $\iint_{S_{2}} d S_{2}$ represents the area of the base, which is $π \cdot 2^{2}$ $= 4 π .$ Then

I_{2} = 4 \iint_{S_{2}} d S_{2} = 4 \cdot 4 π = 16 π .

Thus, the full value of the initial surface integral is

I = I_{1} + I_{2} = 8 \sqrt{2} π + 16 π = 8 π (\sqrt{2} + 2) .