A Worked Upward Flux Problem on a Curved Graph

Setting Up an Upward Flux Calculation

Flux measures how much a vector field passes through an oriented surface. For a curved graph, the calculation depends on three choices made in order: identify the region beneath the graph, represent the graph with a parametrization, and choose the area vector that points upward. Once these are correct, the flux becomes a double integral over the parameter region.

The word upward specifies the orientation: the normal direction must have a positive vertical component. It does not mean that the flux itself must be positive. A field can point partly or mostly downward through an upward-oriented surface, producing negative signed flux.

Definition: Let $S$ be an oriented surface and let $\mathbf{F}$ be a continuous vector field on $S$. The flux of $\mathbf{F}$ across $S$ is $\iint_S \mathbf{F}\cdot\mathbf{n}\,dS$, where $\mathbf{n}$ is the chosen unit normal. Its sign records whether the field points, on balance, with or against that orientation.

The Upward Area Vector for a Graph

Suppose the surface is the graph $z=g(x,y)$ over a region $D$ in the $xy$-plane, with $g$ continuously differentiable. A parametrization is $\mathbf{r}(x,y)=(x,y,g(x,y))$, for $(x,y)\in D$. The partial derivatives are $\mathbf{r}_x=(1,0,g_x)$ and $\mathbf{r}_y=(0,1,g_y)$. Their cross product is $\mathbf{r}_x\times\mathbf{r}_y=(-g_x,-g_y,1)$, whose vertical component is positive. This is therefore the upward oriented area vector per unit area in the parameter plane.

The cross product already includes the surface-area scaling. Consequently, the graph flux can be evaluated directly with this vector; there is no need to multiply by another surface-area factor.

Upward graph formula: If $g$ is continuously differentiable on $D$ and $\mathbf{F}$ is continuous on the graph, then $$ \iint_S \mathbf{F}\cdot\mathbf{n}\,dS = \iint_D \mathbf{F}(x,y,g(x,y))\cdot(-g_x,-g_y,1)\,dA. $$ Here $dA=dx\,dy$, and the vector field is evaluated at points on the surface.

Why the Formula Works

For a parametrized surface, a small rectangle in the parameter plane with side lengths $dx$ and $dy$ maps approximately to a parallelogram on the surface. Its oriented area vector is approximately $(\mathbf{r}_x\times\mathbf{r}_y)\,dx\,dy$. The flux through that small patch is therefore approximately the field at the patch dotted with this area vector. Adding the contributions over the domain gives the parametrized flux integral.

For a graph, the cross product calculation gives precisely $(-g_x,-g_y,1)$. Thus, for an upward orientation,

\mathbf{F}\cdot\mathbf{n}\,dS = \mathbf{F}(\mathbf{r}(x,y))\cdot (\mathbf{r}_x\times\mathbf{r}_y)\,dx\,dy.

This argument also explains why reversing the order of the cross product reverses the orientation: $\mathbf{r}_y\times\mathbf{r}_x=-(\mathbf{r}_x\times\mathbf{r}_y)$, so it changes the sign of every flux contribution.

Worked Example on a Circular Domain

Worked Example: Upward Flux Across a Paraboloid Patch

Find the upward flux of $\mathbf{F}(x,y,z)=(xz,yz,z^2+1)$ across the graph $z=x^2+y^2$ above the unit disk $D=\{(x,y):x^2+y^2\leq 1\}$.

1. Specify the graph and parameter domain. Here $g(x,y)=x^2+y^2$, and the parameter domain is the unit disk. The partial derivatives are $g_x=2x$ and $g_y=2y$, so the upward area vector is $(-2x,-2y,1)$.

2. Evaluate the field on the graph. Set $r^2=x^2+y^2$, so $z=r^2$ on the surface. Then $\mathbf{F}(x,y,g(x,y))=(xr^2,yr^2,r^4+1)$.

3. Form the scalar integrand.

\begin{aligned} \mathbf{F}(x,y,g(x,y))\cdot(-2x,-2y,1) &=-2x^2r^2-2y^2r^2+r^4+1\\ &=-2r^4+r^4+1\\ &=1-r^4. \end{aligned}

4. Integrate over the disk. Use polar coordinates, where $r$ ranges from $0$ to $1$, the angle ranges from $0$ to $2\pi$, and $dA=r\,dr\,d\theta$.

\begin{aligned} \iint_D(1-r^4)\,dA &=\int_0^{2\pi}\int_0^1(1-r^4)r\,dr\,d\theta\\ &=\int_0^{2\pi}\left[\frac{r^2}{2}-\frac{r^6}{6}\right]_0^1d\theta\\ &=\int_0^{2\pi}\left(\frac12-\frac16\right)d\theta\\ &=\int_0^{2\pi}\frac13\,d\theta =\frac{2\pi}{3}. \end{aligned}

Check. On the disk, $0\leq r\leq1$, so $1-r^4\geq0$. The upward-oriented flux is therefore nonnegative, consistent with the positive result $2\pi/3$.

Worked Example on a Rectangle

Worked Example: A Negative Upward Flux

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

The graph derivatives are $g_x=2x$ and $g_y=2y$, so the upward area vector is $(-2x,-2y,1)$. The field does not depend on $z$, and its dot product with that vector is $\mathbf{F}\cdot(-2x,-2y,1)=-2x^2-2y^2+1$. Thus the flux is

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

Check. The total contribution from the constant term is $1$. The $x$-term contributes $-2\int_0^1x^2\,dx=-2/3$, and the $y$-term contributes $-2\int_0^1y^2\,dy=-2/3$. Their sum is $1-2/3-2/3=-1/3$, matching the iterated integral. A negative answer is possible even though the chosen normal points upward: the field has a net component against that orientation over the patch.

Worked Example with a Different Curved Graph

Worked Example: A Smaller Rectangular Parameter Region

Find the upward flux of $\mathbf{F}(x,y,z)=(x,y,1)$ across $z=x^2+2y^2$ over $D=[0,1]\times[0,\tfrac14]$.

Here $g_x=2x$ and $g_y=4y$, so the upward area vector is $(-2x,-4y,1)$. The dot product with the field is $1-2x^2-4y^2$. Integrate each term over the stated rectangle:

\begin{aligned} \iint_D(1-2x^2-4y^2)\,dA &=\int_0^1\int_0^{1/4}(1-2x^2-4y^2)\,dy\,dx\\ &=\frac14-\frac23\left(\frac14\right)-4\left(\frac{1}{192}\right)\\ &=\frac14-\frac16-\frac{1}{48}\\ &=\frac{12}{48}-\frac{8}{48}-\frac{1}{48} =\frac{1}{16}. \end{aligned}

In the second line, the constant term integrates to the rectangle's area $1\cdot\tfrac14=\tfrac14$; the $x^2$ term contributes $-2(\int_0^1x^2\,dx)(\int_0^{1/4}dy)=-\tfrac16$; and the $y^2$ term contributes $-4(\int_0^1dx)(\int_0^{1/4}y^2\,dy)=-\tfrac{1}{48}$. These contributions sum to $1/16$.

Check. The vertical component of the area vector is $1$, so the constant vertical component of the field contributes exactly the projected area $1/4$. The horizontal components subtract $1/6+1/48=3/16$. Hence the net flux is $1/4-3/16=1/16$, consistent with the integral.

A Common Orientation Error

A frequent mistake is to reverse the cross product while still calling the result upward. In the first example, using $\mathbf{r}_y\times\mathbf{r}_x=(2x,2y,-1)$ instead of $\mathbf{r}_x\times\mathbf{r}_y=(-2x,-2y,1)$ would produce

\mathbf{F}(x,y,g(x,y))\cdot(2x,2y,-1) =-(1-r^4).

The resulting integral would be $-2\pi/3$, the negative of the upward flux. That calculation is valid for the downward orientation, but it does not answer an upward-flux question. Check the vertical component of the area vector before integrating: it must be positive for an upward graph orientation.

A related safeguard is to use the area vector exactly once. The vector $(-g_x,-g_y,1)$ already combines the unit normal and the surface-area scaling. Replacing it by a unit normal and then integrating against $dx\,dy$ alone omits the surface-area factor.

Organizing a Complete Solution

For a specified upward graph-flux problem, keep the setup and calculation in a fixed order:

Identify the graph and its domain.
Write $z=g(x,y)$ and specify exactly which $(x,y)$ values belong to $D$.

Compute the upward area vector.
Find $g_x$ and $g_y$, then use $(-g_x,-g_y,1)$. Its positive third component checks the orientation.

Evaluate and dot the field.
Substitute $z=g(x,y)$ into every component of $\mathbf{F}$, then take the dot product with the area vector.

Integrate and interpret.
Use bounds that describe $D$, evaluate the resulting double integral, and check whether the sign is compatible with the integrand.

The sign is a statement about the field relative to the selected orientation, not a test of whether the orientation itself was chosen correctly. A negative flux can be the correct answer for an upward-oriented surface. The orientation check comes from the area vector; the sign check comes from the field's dot product with that vector over the domain.

Check Your Understanding

For each question, distinguish the graph geometry, the orientation, and the signed integral.

For $z=3x^2-y$, what is the upward oriented area vector?
Why does reversing the order in $\mathbf{r}_x\times\mathbf{r}_y$ change the flux sign?
For $z=x^2+y^2$, what is the upward flux integrand for $\mathbf{F}=(0,0,1)$?
Can an upward-oriented surface have negative flux? Explain what that sign means.
Why should a graph-flux calculation not multiply by a separate surface-area factor after using $(-g_x,-g_y,1)$?