Upward Flux Across Graphs Over Rectangles

From Surface Geometry to a Rectangular Integral

A graph surface can be curved, and its upward normal can vary from point to point. Nevertheless, when the graph lies above a rectangle in the $xy$-plane, its flux integral becomes an ordinary double integral with fixed rectangular bounds. The key is to account for both the vector field and the changing slope of the graph.

This method applies whether the field is constant or polynomial, and whether the graph’s partial derivatives are constant or vary across the region. In each case, the calculation has the same structure: find the upward-oriented area element, evaluate the field on the graph, form a scalar integrand, and integrate over the rectangle.

Definition: Let $D=[a,b]\times[c,d]$ be a rectangle, let $g(x,y)$ have continuous first partial derivatives on $D$, and let $\mathbf{F}(x,y,z)=\langle P(x,y,z),Q(x,y,z),R(x,y,z)\rangle$ be continuous on the graph $z=g(x,y)$. The upward flux across that graph is the surface integral of $\mathbf{F}$ in the upward orientation.

The Upward Graph Flux Formula

Parametrize the graph by $\mathbf{r}(x,y)=\langle x,y,g(x,y)\rangle$. Its tangent vectors in the coordinate directions are $\mathbf{r}_x=\langle 1,0,g_x\rangle$ and $\mathbf{r}_y=\langle 0,1,g_y\rangle$. Their cross product points upward because its $z$-component is positive.

\mathbf{r}_x\times\mathbf{r}_y = \langle -g_x,-g_y,1\rangle.

The surface-flux formula pairs the field with this oriented area element. Consequently, the field must be evaluated at points on the graph, not at an unrelated value of $z$.

Formula: If the stated continuity and differentiability conditions hold, the upward flux through $z=g(x,y)$ above $D=[a,b]\times[c,d]$ is $$ \iint_D \mathbf{F}(x,y,g(x,y))\cdot\langle -g_x(x,y),-g_y(x,y),1\rangle\,dA = \int_a^b\int_c^d \left[-P(x,y,g)g_x-Q(x,y,g)g_y+R(x,y,g)\right]\,dy\,dx. $$ Here $g$ in the field components denotes $g(x,y)$. The iterated integral may also be taken in the opposite order because the scalar integrand is continuous on the rectangle.

The dot product explains the three terms. The first two account for the horizontal components of the field meeting a tilted surface; the last accounts for its vertical component. If the field is continuous and the graph is continuously differentiable on the rectangle, the integrand is continuous, so the double integral is well-defined. The orientation is upward specifically because the normal’s $z$-component is $1>0$.

A Reliable Calculation Sequence

Record the rectangle and graph.
Write the bounds and identify $g(x,y)$ before differentiating.

Find both partial derivatives.
Compute $g_x$ and $g_y$, retaining their dependence on $x$ and $y$.

Evaluate the field on the graph.
Substitute $z=g(x,y)$ into $P,Q,$ and $R$.

Form and integrate the scalar integrand.
Use $-Pg_x-Qg_y+R$ with the rectangle’s bounds, then check the result.

This sequence is useful because it separates the geometric contribution of the graph’s slope from the algebra of the field. Keeping the steps visible also makes sign errors easier to locate.

Worked Example: A Paraboloid Above a Unit Square

Find the upward flux of $\mathbf{F}(x,y,z)=\langle x,y,1\rangle$ through $z=x^2+y^2$ over $D=[0,1]\times[0,1]$.

First compute the graph derivatives:

g_x=2x,\qquad g_y=2y.

The components of the field do not depend on $z$. The flux integrand is therefore

$$ -x(2x)-y(2y)+1 = 1-2x^2-2y^2. $$

Integrate over the rectangle, integrating in $y$ first:

\begin{aligned} \Phi &=\int_0^1\int_0^1(1-2x^2-2y^2)\,dy\,dx\\ &=\int_0^1\left[(1-2x^2)y-\frac{2}{3}y^3\right]_0^1\,dx\\ &=\int_0^1\left(\frac13-2x^2\right)\,dx\\ &=\left[\frac{x}{3}-\frac{2x^3}{3}\right]_0^1 =\frac13-\frac23 =-\frac13. \end{aligned}

Check the computation by separating the three terms over the unit square: the constant contributes $1$, the $x^2$ term contributes $-2\int_0^1x^2\,dx=-2/3$, and the $y^2$ term contributes $-2\int_0^1y^2\,dy=-2/3$. Their sum is $1-2/3-2/3=-1/3$, as found. The negative flux means the field’s net component in the upward normal direction is negative over this patch; it does not mean the chosen orientation was downward.

Worked Example: Variable Slopes and a Polynomial Field

Find the upward flux of $\mathbf{F}(x,y,z)=\langle x,y,z+1\rangle$ through $z=x^2y$ above $D=[0,1]\times[0,2]$.

The slope varies with both coordinates:

g_x=2xy,\qquad g_y=x^2.

On the graph, the field becomes $\langle x,y,x^2y+1\rangle$. Thus the integrand is

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

Now integrate over the stated rectangle:

\begin{aligned} \Phi &=\int_0^1\int_0^2(1-2x^2y)\,dy\,dx\\ &=\int_0^1\left[y-x^2y^2\right]_0^2\,dx\\ &=\int_0^1(2-4x^2)\,dx\\ &=\left[2x-\frac{4x^3}{3}\right]_0^1 =2-\frac43 =\frac23. \end{aligned}

For a check, integrate the simplified integrand as a product over the rectangle: the constant contributes $1\cdot2=2$, while $2x^2y$ contributes $2(1/3)(2)=4/3$. Hence the flux is $2-4/3=2/3$. The cancellation in the integrand depended on evaluating the vertical field component at $z=x^2y$.

Using the Rectangle’s Symmetry

Rectangular bounds can make symmetry visible. An odd function of $x$ integrates to zero over bounds symmetric about $x=0$. This can simplify a flux calculation, but symmetry should only be used after the full flux integrand has been formed: individual terms may have different parity.

Worked Example: A Symmetric Rectangle and a Negative Flux

Find the upward flux of $\mathbf{F}(x,y,z)=\langle y,x,z\rangle$ through $z=xy^2$ above $D=[-1,1]\times[0,1]$.

The derivatives and field evaluated on the graph are

g_x=y^2,\qquad g_y=2xy,\qquad \mathbf{F}(x,y,g)=\langle y,x,xy^2\rangle.

Therefore the scalar integrand is

$$ -y(y^2)-x(2xy)+xy^2 =-y^3-2x^2y+xy^2. $$

Integrate term by term on the rectangle:

\begin{aligned} \Phi &=\int_{-1}^{1}\int_0^1(-y^3-2x^2y+xy^2)\,dy\,dx\\ &=\int_{-1}^{1}\left(-\frac14-x^2+\frac{x}{3}\right)\,dx\\ &=\left[-\frac{x}{4}-\frac{x^3}{3}+\frac{x^2}{6}\right]_{-1}^{1}\\ &=\left(-\frac14-\frac13+\frac16\right) -\left(\frac14+\frac13+\frac16\right)\\ &=-\frac{5}{12}-\frac{3}{4} =-\frac{14}{12} =-\frac76. \end{aligned}

Check by parity: the $xy^2$ term is odd in $x$, so its integral over $[-1,1]$ is zero. The $-y^3$ term contributes $-2(1/4)=-1/2$, and $-2x^2y$ contributes $-2(2/3)(1/2)=-2/3$. Thus $-1/2-2/3=-7/6$, agreeing with the iterated integral.

Common Pitfall: Dropping a Slope Term

A frequent error is to treat the upward normal as vertical and integrate only the vertical component $R$. That works for a horizontal graph, where $g_x=g_y=0$, but not in general. The horizontal components $P$ and $Q$ contribute whenever the graph is tilted.

Worked Example: Correcting an Omitted-Term Error

For the first example, the vertical component was $R=1$. An incorrect calculation would therefore claim

\Phi_{\text{wrong}}=\int_0^1\int_0^1 1\,dy\,dx=1.

This omits both slope contributions. For $g=x^2+y^2$, $g_x=2x$ and $g_y=2y$, while $P=x$ and $Q=y$. The correct integrand is

$$ -Pg_x-Qg_y+R=-2x^2-2y^2+1. $$

As computed above, its integral is $1-2/3-2/3=-1/3$, not $1$. Substituting the values back into the required dot product confirms the correction:

\langle x,y,1\rangle\cdot\langle-2x,-2y,1\rangle =-2x^2-2y^2+1.

A different sign error comes from replacing the upward normal $\langle-g_x,-g_y,1\rangle$ with its negative. That changes the orientation and reverses the flux. Keep the positive third component when the problem specifies upward orientation.

Why the Formula Works

The graph parametrization maps each point $(x,y)$ in the rectangle to the surface point $(x,y,g(x,y))$. The cross product of its coordinate tangent vectors gives the oriented area vector per unit area in the parameter plane. Pairing the field at that surface point with this vector measures the field component crossing the surface, weighted by the surface’s projected area element. Integrating over the rectangle adds those contributions across the entire patch.

This reasoning also explains why the formula has no factor $\sqrt{1+g_x^2+g_y^2}$ by itself. That factor belongs to the scalar surface-area element. For flux, the normal direction and area magnitude are combined in the cross product; the resulting vector area element is $\langle-g_x,-g_y,1\rangle\,dA$. Introducing the scalar area factor separately without the corresponding unit normal would count the area incorrectly.

Check Your Understanding

For each question, use the upward orientation and keep the graph’s rectangular bounds explicit.

For a graph $z=g(x,y)$, what is the $z$-component of the upward-oriented area vector?
What substitution must be made in the components of $\mathbf{F}(x,y,z)$ before forming the graph flux integrand?
For $g(x,y)=x^2+y^2$, compute $g_x$ and $g_y$.
Why is integrating only the vertical field component generally insufficient for an upward flux calculation?
What change in flux results if the orientation is reversed while the surface and field remain fixed?