]> 31.3 Determining the Integrand

## 31.3 Determining the Integrand

A flux integral is one of the general form

$∬ S W ⟶ · n ^ d S$

Suppose now we have a parametric representation for $S$ , which means we have equations $x = x ( u , v ) , y = y ( u , v )$ and $z = z ( u , v )$ , which define our surface.

We can reduce this integral to a sequence of ordinary integrals over $u$ and $v$ by expressing $( W ⟶ · n ^ ) d S$ as an explicit function of $u$ and $v$ multiplied by $d u d v$ and determining an order and limits on the resulting integrals.

The first step in this procedure is straightforward:

Form the vectors

$∂ x ∂ u i ^ + ∂ y ∂ u j ^ + ∂ z ∂ u k ^$

which we shall call

$∂ P ⟶ ∂ u$

and similarly

$∂ x ∂ v i ^ + ∂ y ∂ v j ^ + ∂ z ∂ v k ^$

which we shall call

$∂ P ⟶ ∂ v$

Then the integrand, integrating over $u$ and $v$ , can be written as the absolute value of the determinant of the matrix with columns given by the components of $W ⟶ , ∂ P ⟶ ∂ u$ , and $∂ P ⟶ ∂ v$ , multiplied by an appropriate sign.

For the reason why, see section 24.1 . (The vector $n ^$ here in the integral is generally meant to represent the outward directed normal with respect to some region and the sign of the integral is the one such that if $W ⟶$ has positive dot product with this outward direction, the result is positive.)

The appropriate sign must be determined separately, but only once per surface, from the context of the original integral.

Thus to within a sign, the integral becomes

$∬ | W ⟶ , ∂ P ⟶ ∂ u , ∂ P ⟶ ∂ v | d u d v$

While this reduction straightforward, the steps necessary to perform it in practice involve a sufficient amount of algebraic manipulation that I for one almost always make at least one error in doing it, and so rarely get the same answer twice.

Fortunately there is a computationally simpler answer in the most common case, in which $u$ and $v$ are actually two of your original variables, say $x$ and $y$ .

In that case the vectors $∂ P ⟶ ∂ u$ , and $∂ P ⟶ ∂ v$ , become $( 1 , 0 , ∂ z ∂ x )$ , and $( 0 , 1 , ∂ z ∂ y )$ and their cross products becomes $( − ∂ z ∂ x , − ∂ z ∂ y , 1 )$ .

The integrand and area element can then be written as

$( W z − W x ∂ z ∂ x − W y ∂ z ∂ y ) d x d y$

which result is much easier to apply. We have assumed here that the sign should be positive in the direction that is upward in the $z$ direction, and the sign must be reversed if that assumption is incorrect.

If your surface is described by an equation $f ( x , y , z ) = 0$ , the corresponding formula with the same sign assumption becomes

$( W ⟶ · ∇ ⟶ f ∂ f / ∂ z ) d x d y$

(When $∂ f ∂ z = 0$ you must integrate over two other variables, say $y$ and $z$ instead of $x$ and $y$ .)

The answers given here give both the integrand and the area element.

The only time you really have to worry about the area element is when you want to change variables and have to determine what the area element $d x d y$ means in terms of other variables $u$ and $v$ .

Here is an example: we want to compute the flux of the vector $( x , y , z )$ outward through the surface defined by the equation

$( x A ) 2 + ( y B ) 2 + ( z C ) 2 − 1 = 0$

(This surface is called an ellipsoid.)

We will integrate over the portion of the surface for which $z > 0$ only and double the result since the lower part of the surface gives the same flux.

Writing this as $f = 0$ we compute $∇ ⟶ f = ( 2 x A 2 , 2 y B 2 , 2 z C 2 )$ and the integrand becomes, with the last formula, $W ⟶ · ∇ ⟶ f ∂ W ⟶ / ∂ z$ which in this case is $C 2 z$ , which is $( 1 − ( x A ) 2 − ( y B ) 2 ) − 1 2$ .

Our integral therefore becomes

$∬ d x d y ( 1 − x 2 A 2 − y 2 B 2 ) 1 2$

over appropriate limits of integration. Those limits are the bounds (which we will discuss below) determined by the area in the $x y$ plane that we are integrating over, and that is, the region in which the denominator in that integral is positive.