Deriving the Upward Graph Flux Formula

From a Graph to an Oriented Area Vector

A surface given by $z=f(x,y)$ is determined by its height above each point $(x,y)$ in a region $D$ of the $xy$-plane. Its geometry is encoded by the partial derivatives $f_x$ and $f_y$: these describe how the height changes in the two coordinate directions.

For flux, however, we need more than the surface’s shape. We also need an orientation. Here we derive the area vector for the orientation that points upward, and use it to obtain a compact formula for flux through the graph.

Setup: Let $D$ be a region in the $xy$-plane, let $f$ have continuous first partial derivatives on an open set containing $D$, and let $\mathbf{F}=(P,Q,R)$ be continuous along the graph. The graph is parametrized by $\mathbf{r}(x,y)=(x,y,f(x,y))$, for $(x,y)\in D$.

Parametrizing the Graph

The parametrization assigns a point on the surface to each point in its domain:

\mathbf{r}(x,y)=(x,y,f(x,y)).

To see how the surface changes as we move in the parameter directions, differentiate with respect to $x$ and $y$. The resulting vectors lie tangent to the surface.

\mathbf{r}_x=(1,0,f_x), \qquad \mathbf{r}_y=(0,1,f_y).

The cross product of these tangent vectors is perpendicular to both. Its order matters: $\mathbf{r}_x\times\mathbf{r}_y$ and $\mathbf{r}_y\times\mathbf{r}_x$ have opposite directions.

\mathbf{r}_x\times\mathbf{r}_y = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1&0&f_x\\ 0&1&f_y \end{vmatrix} = (-f_x,-f_y,1).

The third component is $1$, which is positive. Thus this cross product points upward, meaning its vertical component is positive. It gives the upward orientation without requiring us to normalize the vector.

Definition: For the graph $z=f(x,y)$ with upward orientation, the oriented area vector is $\mathbf{r}_x\times\mathbf{r}_y\,dx\,dy=(-f_x,-f_y,1)\,dx\,dy$.

Deriving the Upward Flux Formula

For a parametrized surface, the flux of $\mathbf{F}$ is obtained by taking its dot product with the oriented area vector and integrating over the parameter domain. Here the field is evaluated at the point $\mathbf{r}(x,y)$, whose coordinates are $(x,y,f(x,y))$. Write its components there as $P(x,y,f(x,y))$, $Q(x,y,f(x,y))$, and $R(x,y,f(x,y))$.

\begin{aligned} \mathbf{F}(\mathbf{r}(x,y))\cdot (\mathbf{r}_x\times\mathbf{r}_y) &=(P,Q,R)\cdot(-f_x,-f_y,1)\\ &=-P f_x-Q f_y+R. \end{aligned}

Consequently, the flux through the graph with upward orientation is the integral over its parameter domain of this dot product.

Theorem: Under the setup above, the upward flux through the graph of $f$ over $D$ is $$ \iint_D \mathbf{F}(x,y,f(x,y))\cdot(-f_x,-f_y,1)\,dA = \iint_D \bigl(-P(x,y,f(x,y))f_x-Q(x,y,f(x,y))f_y+R(x,y,f(x,y))\bigr)\,dA, $$ provided the displayed integrand is integrable over $D$. Here $dA=dx\,dy$ denotes area in the parameter plane.

Why the Derivation Works

The cross product does two jobs at once. Its direction is normal to the surface, and its magnitude accounts for how much surface area corresponds to a small parameter-plane area. Therefore it is the oriented area vector needed in the definition of parametrized flux.

In this case, the magnitude is

\left\lVert\mathbf{r}_x\times\mathbf{r}_y\right\rVert = \sqrt{f_x^2+f_y^2+1}.

The upward unit normal and scalar surface-area element can therefore be written as

\mathbf{n} = \frac{(-f_x,-f_y,1)}{\sqrt{1+f_x^2+f_y^2}}, \qquad dS=\sqrt{1+f_x^2+f_y^2}\,dA.

Multiplying these expressions cancels the square-root factor: $\mathbf{n}\,dS=(-f_x,-f_y,1)\,dA$. That cancellation explains why the graph-flux formula uses the oriented area vector directly, without a square root in its integrand.

Worked Examples

Worked Example: Find the Upward Area Vector

Let $f(x,y)=x^2+3y$. Find the upward oriented area vector at a general parameter point, then evaluate it at $(1,2)$.

First compute the partial derivatives:

f_x=2x,\qquad f_y=3.

Substitute them into the graph area-vector formula:

\mathbf{r}_x\times\mathbf{r}_y=(-2x,-3,1).

At $(x,y)=(1,2)$, the point on the graph has height $f(1,2)=1+6=7$, and the upward area-vector components are

$$ (-2(1),-3,1)=(-2,-3,1). $$

Check: The third component is $1>0$, so the vector points upward, as required. Its dot product with each tangent vector is zero: $(-2,-3,1)\cdot(1,0,2)=-2+2=0$, and $(-2,-3,1)\cdot(0,1,3)=-3+3=0$.

Worked Example: Obtain a Flux Integrand

Let $z=xy$, and let $\mathbf{F}(x,y,z)=(z,x,y)$. Find the upward flux integrand in the parameter plane, and evaluate it at $(1,2)$.

For the graph $f(x,y)=xy$, its partial derivatives are

f_x=y,\qquad f_y=x.

The field evaluated on the graph is $\mathbf{F}(x,y,xy)=(xy,x,y)$. Its dot product with the upward oriented area vector gives

\begin{aligned} \mathbf{F}(x,y,xy)\cdot(-y,-x,1) &=(xy,x,y)\cdot(-y,-x,1)\\ &=-xy^2-x^2+y. \end{aligned}

At $(1,2)$, the graph height is $z=1\cdot2=2$, so the field value is $(2,1,2)$ and the area-vector components are $(-2,-1,1)$. The integrand is

(2,1,2)\cdot(-2,-1,1)=-4-1+2=-3.

Check: Substituting $x=1$ and $y=2$ into the derived expression gives $-1(2^2)-1^2+2=-4-1+2=-3$, the same value.

Worked Example: Compute Flux over a Triangular Parameter Domain

Let $D=\{(x,y):0\le x\le1,\ 0\le y\le1-x\}$, let the graph be $z=x+y$, and let $\mathbf{F}(x,y,z)=(z,x,y)$. Compute the upward flux using the graph formula.

Here $f_x=1$, $f_y=1$, and the field on the graph is $\mathbf{F}(x,y,x+y)=(x+y,x,y)$. Therefore the flux integrand is

(x+y,x,y)\cdot(-1,-1,1)=-(x+y)-x+y=-2x.

Integrate over the stated bounds:

\begin{aligned} \text{Flux} &=\int_0^1\int_0^{1-x}(-2x)\,dy\,dx\\ &=\int_0^1[-2xy]_{y=0}^{y=1-x}\,dx\\ &=\int_0^1 -2x(1-x)\,dx\\ &=\int_0^1(-2x+2x^2)\,dx\\ &=[-x^2+\tfrac{2}{3}x^3]_0^1\\ &=-1+\tfrac{2}{3}=-\tfrac{1}{3}. \end{aligned}

Check: Directly, $\mathbf{r}_x=(1,0,1)$, $\mathbf{r}_y=(0,1,1)$, and $\mathbf{r}_x\times\mathbf{r}_y=(-1,-1,1)$, so this is the upward orientation. The field dot this cross product is $-(x+y)-x+y=-2x$, which reproduces the integrand and the computed flux $-\tfrac13$.

Common Pitfall: Dropping a Sign or a Factor

A frequent error is to change the order of the cross product without changing the orientation. For example, with $f(x,y)=xy$, the upward cross product is $(-y,-x,1)$. Reversing the order gives $(y,x,-1)$, whose third component is negative, so it points downward.

Using that reversed vector while claiming to calculate upward flux changes the sign. For the field in the previous example, the incorrect calculation would use

(xy,x,y)\cdot(y,x,-1)=xy^2+x^2-y,

which is the negative of the correct integrand $-xy^2-x^2+y$. The correction is to keep the parameter order $\mathbf{r}_x\times\mathbf{r}_y$ for the upward graph orientation, or to verify the direction by checking that the resulting vector has positive third component.

Another error is to use the unit normal alone as the area vector. The unit normal has length $1$, but flux also needs the surface-area factor. If using a unit normal, include $dS$; if using the unnormalized cross product, integrate against $dA$. The identity $\mathbf{n}\,dS=(-f_x,-f_y,1)\,dA$ makes the two equivalent approaches explicit.

Why the Formula Is Useful

The formula converts a flux calculation over a graph into a two-variable integral over its parameter domain. It avoids separately computing a unit normal and a surface-area factor, while keeping the orientation visible in the components $(-f_x,-f_y,1)$. It also shows exactly how each component of the field contributes: the horizontal components are weighted by the negative slopes of the graph, while the vertical component enters with a positive sign for upward orientation.

Before using the formula, check that the surface really is represented as $z=f(x,y)$ on the domain in question, that the required partial derivatives exist, and that the field is evaluated at $z=f(x,y)$. Then check the orientation and the signs in the dot product. These checks prevent many errors before any integration begins.

Check Your Understanding

Use the graph parametrization and upward orientation to answer these questions.

For $z=f(x,y)$, what are $\mathbf{r}_x$ and $\mathbf{r}_y$?
Compute $\mathbf{r}_x\times\mathbf{r}_y$, and explain how its components show that it points upward.
For $\mathbf{F}=(P,Q,R)$, write the upward flux integrand on the graph, making clear where the field is evaluated.
Why is it incorrect to use the unit normal without also including the scalar surface-area element?
If the cross-product order is reversed, what happens to its direction and to the signed flux?