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Flux Integrals and Orientation · Tutorial 886 of 999

Downward Flux Across Graph Surfaces

Use the downward oriented area vector for a graph \(z=g(x,y)\) to compute flux and verify its sign.

Advanced 8 min read

What You'll Learn

  • Identify the downward orientation of a graph over a region in the xy-plane.
  • Derive the downward oriented area vector from a graph parametrization.
  • Apply the downward graph flux formula to continuous vector fields.
  • Compute flux over a disk, rectangle, and triangular region.
  • Check the orientation sign using a constant vertical field.
  • Recognize and correct the error of using the upward area vector.

Why Downward Orientation Changes the Flux

A graph surface \(z=g(x,y)\) has two possible orientations: one points generally upward, and the other points generally downward. The surface itself does not change, but the oriented area vector reverses when the chosen orientation reverses. Since flux measures the component of a vector field in the chosen normal direction, reversing orientation changes the sign of the flux.

For a graph over a region \(D\) in the \(xy\)-plane, downward orientation means that the normal vector has a negative \(z\)-component. The key computational step is to use the downward oriented area vector, not the upward one.

Definition: A graph \(z=g(x,y)\) over \(D\) is downward oriented when its chosen normal has negative \(z\)-component at every point. Its downward flux is the integral of the vector field’s normal component over that oriented surface.

The Downward Graph Flux Formula

Suppose \(D\) is a bounded region with piecewise smooth boundary, \(g\) has continuous first partial derivatives on a region containing \(D\), and \(\mathbf{F}=(P,Q,R)\) is continuous on the graph. The graph can be parametrized by \(\mathbf{r}(x,y)=(x,y,g(x,y))\). Its downward oriented area vector, expressed over the projection \(D\), is \((g_x,g_y,-1)\,dA\). Therefore the downward flux is:

$$ \iint_S \mathbf{F}\cdot d\mathbf{S} = \iint_D \left[ P(x,y,g(x,y))g_x(x,y) + Q(x,y,g(x,y))g_y(x,y) - R(x,y,g(x,y)) \right]\,dA. $$

Here \(dA=dx\,dy\) is area in the projection plane. Each component of \(\mathbf{F}\) is evaluated at the corresponding point on the surface, \((x,y,g(x,y))\). The formula uses an oriented area vector; it does not require multiplying by the scalar surface-area factor \(\sqrt{1+g_x^2+g_y^2}\) separately.

Why the Formula Works

The parameter directions of the graph are \(\mathbf{r}_x=(1,0,g_x)\) and \(\mathbf{r}_y=(0,1,g_y)\). Their cross product gives an oriented area vector for the parametrization’s order:

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

This vector points upward because its \(z\)-component is \(1\). To obtain the downward orientation, multiply the entire vector by \(-1\). The resulting vector is \((g_x,g_y,-1)\), whose \(z\)-component is \(-1\). Dotting it with \(\mathbf{F}(x,y,g(x,y))\) gives the integrand in the formula.

$$ \mathbf{F}(x,y,g(x,y))\cdot(g_x,g_y,-1) = P(x,y,g(x,y))g_x + Q(x,y,g(x,y))g_y - R(x,y,g(x,y)). $$

Integrating this dot product over \(D\) sums the field’s signed normal component across the graph. The sign is part of the orientation: components in the downward direction contribute positively, while components in the upward direction contribute negatively.

Orientation check: For a downward oriented graph, the third component of the oriented area vector must be negative. If it is \(+1\), the upward orientation has been used instead.

Worked Example: A Constant Downward Field Over a Disk

Worked Example: A Constant Downward Field Over a Disk

Find the downward flux of \(\mathbf{F}=(0,0,-4)\) through \(z=x^2+y^2\) over the unit disk \(D=\{(x,y):x^2+y^2\leq 1\}\).

First compute the derivatives of the graph function \(g(x,y)=x^2+y^2\):

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

For this field, \(P=0\), \(Q=0\), and \(R=-4\). The downward flux integrand is therefore

$$ P g_x+Q g_y-R = 0(2x)+0(2y)-(-4) = 4. $$

The integrand is constant, so its integral over the unit disk is \(4\) times the disk’s area:

$$ \iint_D 4\,dA = 4\pi(1)^2 = 4\pi. $$

Thus the downward flux is \(4\pi\). As a sign check, the downward oriented area vector is \((2x,2y,-1)\), and its dot product with \((0,0,-4)\) is \(4\) at every point. A downward field has a positive component in the downward direction, as this positive answer indicates. The result also equals the field’s downward magnitude, \(4\), multiplied by the area of the projected disk, \(\pi\).

A Constant Vertical Field as a General Sign Test

The preceding example illustrates a useful check that applies to any graph over \(D\). Let the field be the constant vertical field \(\mathbf{F}=(0,0,c)\), where \(c\) is a real constant. Since \(P=Q=0\), the downward formula reduces to:

$$ \iint_D -c\,dA = -c\,\operatorname{Area}(D). $$

If \(c<0\), the field points downward and the downward flux is positive. If \(c>0\), the field points upward and the downward flux is negative. If \(c=0\), the flux is zero. The graph’s slope does not affect this particular flux: a vertical field’s dot product with the downward oriented area vector depends only on its vertical component, \(-1\).

Worked Example: A Sloping Graph Over a Rectangle

Worked Example: A Sloping Graph Over a Rectangle

Compute the downward flux of \(\mathbf{F}=(z,2x,1)\) through \(z=x^2+y\) over \(D=[0,1]\times[0,1]\).

For \(g(x,y)=x^2+y\), the partial derivatives are \(g_x=2x\) and \(g_y=1\). On the graph, \(P=z=x^2+y\), \(Q=2x\), and \(R=1\). Substituting into the downward integrand gives

$$ P g_x+Q g_y-R = (x^2+y)(2x)+(2x)(1)-1 = 2x^3+2xy+2x-1. $$

Integrate first with respect to \(y\), from \(0\) to \(1\):

$$ \int_0^1(2x^3+2xy+2x-1)\,dy = 2x^3+ x+2x-1 = 2x^3+3x-1. $$

Now integrate with respect to \(x\), from \(0\) to \(1\):

$$ \int_0^1(2x^3+3x-1)\,dx = \left[\frac{x^4}{2}+\frac{3x^2}{2}-x\right]_0^1 = \frac12+\frac32-1 = 1. $$

The downward flux is \(1\). To check the integration, integrate the four terms of the original integrand separately over the unit square. Their contributions are \(1/2\), \(1/2\), \(1\), and \(-1\), respectively; their sum is \(1/2+1/2+1-1=1\), agreeing with the iterated integral.

Worked Example: A Horizontal Field Over a Triangular Region

Worked Example: A Horizontal Field Over a Triangular Region

Find the downward flux of \(\mathbf{F}=(1,0,0)\) through \(z=xy\) over the triangle \(D=\{(x,y):0\leq x\leq 1,\;0\leq y\leq 1-x\}\).

For \(g(x,y)=xy\), we have \(g_x=y\) and \(g_y=x\). Here \(P=1\), \(Q=0\), and \(R=0\), so the downward integrand is

$$ P g_x+Q g_y-R = 1(y)+0(x)-0 = y. $$

Using the given bounds, integrate first with respect to \(y\):

$$ \int_0^1\int_0^{1-x} y\,dy\,dx = \int_0^1\left[\frac{y^2}{2}\right]_0^{1-x}dx = \frac12\int_0^1(1-x)^2\,dx. $$

Expanding and integrating gives

$$ \frac12\int_0^1(1-2x+x^2)\,dx = \frac12\left[x-x^2+\frac{x^3}{3}\right]_0^1 = \frac12\left(1-1+\frac13\right) = \frac16. $$

The downward flux is \(1/6\). The result is positive because the downward oriented area vector \((y,x,-1)\) has first component \(y\), which has nonnegative dot product with the field \((1,0,0)\) over this triangle. The direct integral also confirms the value: the average value of \(y\) over a triangle of area \(1/2\) is \(1/3\), giving \((1/3)(1/2)=1/6\).

A Common Orientation Error

A frequent mistake is to calculate \(\mathbf{r}_x\times\mathbf{r}_y=(-g_x,-g_y,1)\) and use it for a downward flux without reversing its sign. That vector is upward oriented. For instance, in the disk example, using the upward vector with \(\mathbf{F}=(0,0,-4)\) gives an integrand of \(-4\) and the incorrect value \(-4\pi\). The downward vector is its negative, so the correct integrand is \(4\) and the correct flux is \(4\pi\).

Correction: Determine the required orientation before taking the dot product. For downward flux through \(z=g(x,y)\), use \((g_x,g_y,-1)\), or equivalently negate the upward vector \((-g_x,-g_y,1)\). Reversing orientation reverses the sign of the entire flux.

Applying the Formula Reliably

For a downward graph-flux calculation, keep the geometry and the integration separate. First identify \(D\) and \(g\), then differentiate \(g\), evaluate the field on the graph, and form the downward dot product. Only after that should you set up the integral using bounds for \(D\). This order makes it easier to catch both orientation errors and incorrect substitutions into the field.

The formula applies when the graph is differentiable enough to have the stated partial derivatives and the field is continuous on the surface, with the region satisfying the stated boundary conditions. For a region described by different bounds, the integrand stays the same; only the integration limits change. The projected area element is always \(dA\), and the downward orientation is always indicated by the negative third component of the oriented area vector.

Check Your Understanding

For each question, pay attention to the orientation as well as the integral.

  1. What is the downward oriented area vector for the graph \(z=g(x,y)\), expressed over its projection?
  2. For a constant field \(\mathbf{F}=(0,0,5)\), is the downward flux positive or negative? Express it in terms of the area of \(D\).
  3. What downward flux formula results for \(\mathbf{F}=(P,Q,R)\) on \(z=g(x,y)\)? At what points should the field components be evaluated?
  4. For \(z=3x-y\) over the unit square, what is the downward flux of \(\mathbf{F}=(0,0,-2)\)?
  5. Why does using \((-g_x,-g_y,1)\) for a downward flux produce a sign error?

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