Surface Integrals of Vector Fields

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We consider a vector field $F (x, y, z)$ and a surface $S,$ which is defined by the position vector

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

Suppose that the functions $x (u, v),$ $y (u, v),$ $z (u, v)$ are continuously differentiable in some domain $D (u, v)$ and the rank of the matrix

[\begin{matrix} \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} & \frac{\partial z}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} & \frac{\partial z}{\partial v} \end{matrix}]

is equal to $2.$

We denote by $n (x, y, z)$ a unit normal vector to the surface $S$ at the point $(x, y, z) .$ If the surface $S$ is smooth and the vector function $n (x, y, z)$ is continuous, there are only two possible choices for the unit normal vector:

n (x, y, z) or - n (x, y, z) .

If the choice of the vector is done, the surface $S$ is called oriented.

If $S$ is a closed surface, by convention, we choose the normal vector to point outward from the surface.

The surface integral of the vector field $F$ over the oriented surface $S$ (or the flux of the vector field $F$ across the surface $S$ ) can be written in one of the following forms:

If the surface $S$ is oriented outward, then
$\iint_{S} F (x, y, z) \cdot d S = \iint_{S} F (x, y, z) \cdot n d S = \iint_{D (u, v)} F (x (u, v), y (u, v), z (u, v)) [\frac{\partial r}{\partial u} \times \frac{\partial r}{\partial v}] d u d v;$
If the surface $S$ is oriented inward, then
$\iint_{S} F (x, y, z) \cdot d S = \iint_{S} F (x, y, z) \cdot n d S = \iint_{D (u, v)} F (x (u, v), y (u, v), z (u, v)) [\frac{\partial r}{\partial v} \times \frac{\partial r}{\partial u}] d u d v .$

Here $d S = n d S$ is called the vector element of the surface. Dot means the scalar product of the appropriate vectors. The partial derivatives in the formulas are calculated in the following way:

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

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

If the surface $S$ is given explicitly 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 of the vector field $F$ over the surface $S$ is defined in one of the following forms:

If the surface $S$ is oriented upward, i.e. the $k$ th component of the normal vector is positive, then
$\iint_{S} F (x, y, z) \cdot d S = \iint_{S} F (x, y, z) \cdot n d S = \iint_{D (x, y)} F (x, y, z) (- \frac{\partial z}{\partial x} i - \frac{\partial z}{\partial y} j + k) d x d y;$
If the surface $S$ is oriented downward, i.e. the $k$ th component of the normal vector is negative, then
$\iint_{S} F (x, y, z) \cdot d S = \iint_{S} F (x, y, z) \cdot n d S = \iint_{D (x, y)} F (x, y, z) (\frac{\partial z}{\partial x} i + \frac{\partial z}{\partial y} j - k) d x d y .$

We can also write the surface integral of vector fields in the coordinate form.

Let $P (x, y, z),$ $Q (x, y, z),$ $R (x, y, z)$ be the components of the vector field $F .$ Suppose that $\cos α,$ $\cos β,$ $\cos γ$ are the angles between the outer unit normal vector $n$ and the $x$ -axis, $y$ -axis, and $z$ -axis, respectively. Then the scalar product $F \cdot n$ is

F \cdot n = F (P, Q, R) n (\cos α, \cos β, \cos γ) = P \cos α + Q \cos β + R \cos γ .

Consequently, the surface integral can be written as

\iint_{S} (F \cdot n) d S = \iint_{S} (P \cos α + Q \cos β + R \cos γ) d S .

As $\cos α \cdot d S = d y d z$ (Figure $1$ ), and, similarly, $\cos β \cdot d S = d z d x,$ $\cos γ \cdot d S = d x d y,$ we obtain the following formula for calculating the surface integral:

\iint_{S} (F \cdot n) d S = \iint_{S} (P \cos α + Q \cos β + R \cos γ) d S = \iint_{S} P d y d z + Q d z d x + R d x d y .

If the surface $S$ is given in parametric form by the vector $r (x (u, v), y (u, v),$ $z (u, v)),$ the latter formula can be written as

\iint_{S} (F \cdot n) d S = \iint_{S} P d y d z + Q d z d x + R d x d y = \iint_{D (u, v)} | \begin{matrix} P & Q & R \\ \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} & \frac{\partial z}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} & \frac{\partial z}{\partial v} \end{matrix} | d u d v,

where the coordinates $(u, v)$ range over some domain $D (u, v) .$

Solved Problems

Click or tap a problem to see the solution.

Example 1

Evaluate the flux of the vector field $F (x, y, z) = (x, - 1, z)$ across the surface $S$ that has downward orientation and is given by the equation

z = x \cos y, 0 \leq x \leq 1, \frac{π}{4} \leq y \leq \frac{π}{3} .

Example 2

Find the flux of the vector field $F (x, y, z) = (y, x, z)$ through the surface $S,$ parameterized by the vector

r (u, v) = (\cos v, \sin v, u), 0 \leq u \leq 2, \frac{π}{2} \leq v \leq π .

Example 1.

Evaluate the flux of the vector field $F (x, y, z) = (x, - 1, z)$ across the surface $S$ that has downward orientation and is given by the equation

z = x \cos y, 0 \leq x \leq 1, \frac{π}{4} \leq y \leq \frac{π}{3} .

Solution.

We apply the formula

\iint_{S} F \cdot d S = \iint_{D (x, y)} F (\frac{\partial z}{\partial x} i + \frac{\partial z}{\partial y} j - k) d x d y .

Since

\frac{\partial z}{\partial x} = \frac{\partial}{\partial x} (x \cos y) = \cos y, \frac{\partial z}{\partial y} = \frac{\partial}{\partial y} (x \cos y) = - x \sin y,

the flux of the vector field can be written as

\iint_{S} F \cdot d S = \iint_{D (x, y)} [x \cdot \cos y + (- 1) \cdot (- x \sin y) + z \cdot (- 1)] d x d y = \iint_{D (x, y)} (x \cos y + x \sin y - x \cos y) d x d y = \iint_{D (x, y)} x \sin y d x d y .

After some algebra we find the answer:

\iint_{S} F \cdot d S = \int_{0}^{1} x d x \int_{\frac{π}{4}}^{\frac{π}{3}} \sin y d y = [{(\frac{x^{2}}{2}) |}_{0}^{1}] \cdot [{(- \cos y) |}_{\frac{π}{4}}^{\frac{π}{3}}] = \frac{1}{2} (- \cos \frac{π}{3} + \cos \frac{π}{4}) = \frac{1}{2} (- \frac{1}{2} + \frac{\sqrt{2}}{2}) = \frac{\sqrt{2} - 1}{4} .

Example 2.

Find the flux of the vector field $F (x, y, z) = (y, x, z)$ through the surface $S,$ parameterized by the vector

r (u, v) = (\cos v, \sin v, u), 0 \leq u \leq 2, \frac{π}{2} \leq v \leq π .

Solution.

First we calculate the partial derivatives:

\frac{\partial r}{\partial u} = (0, 0, 1), \frac{\partial r}{\partial v} = (- \sin v, \cos v, 0) .

It follows that

\frac{\partial r}{\partial u} \times \frac{\partial r}{\partial v} = | \begin{matrix} i & j & k \\ 0 & 0 & 1 \\ - \sin v & \cos v & 0 \end{matrix} | = - \cos v \cdot i - \sin v \cdot j .

Hence, the vector area element is

d S = [\frac{\partial r}{\partial u} \times \frac{\partial r}{\partial v}] d u d v = (- \cos v, - \sin v, 0) d u d v .

As $x = \cos v,$ $y = \sin v$ and $z = v,$ the vector field $F$ can be represented in the following form:

F (r, u, v) = (\sin v, \cos v, u) .

Then the original surface integral across $S$ is

\iint_{S} F \cdot d S = \iint_{D (u, v)} [\sin v \cdot (- \cos v) + \cos v \cdot (- \sin v) + 0] d u d v = \iint_{D (u, v)} (- 2 \sin v \cos v) d u d v = - \int_{0}^{2} d u \int_{\frac{π}{2}}^{π} \sin 2 v d v = - 2 \cdot [{(- \frac{\cos 2 v}{2}) |}_{\frac{π}{2}}^{π}] = \cos 2 π - \cos π = 2.