Flux Through Planar Patches

Flux Across a Planar Patch

A planar patch has a useful feature: if it is parametrized affinely, its two tangent vectors do not change from point to point. Their cross product—the oriented area vector used in a flux integral—is therefore constant. This turns the surface integral into a double integral over the parameter domain, without requiring a varying normal vector.

The procedure is compact, but keeping the orientation and area factor straight matters. The cross product is not merely a direction: its magnitude carries the area scaling of the parametrization. The examples below show how that vector enters the flux calculation and how the parameter domain determines the integral’s bounds.

Definition: An affine parametrization of a planar patch has the form $\mathbf{r}(u,v)=\mathbf{p}+u\mathbf{a}+v\mathbf{b}$, where $\mathbf{p},\mathbf{a},\mathbf{b}$ are fixed vectors and $\mathbf{a},\mathbf{b}$ are linearly independent. The patch is the image of a parameter domain $D$ under $\mathbf{r}$. Its orientation is specified by the order $u,v$, giving the oriented area vector $\mathbf{a}\times\mathbf{b}$.

Linear independence of $\mathbf{a}$ and $\mathbf{b}$ ensures that the image is genuinely two-dimensional. The parameter domain $D$ is typically a bounded region whose boundary is piecewise smooth, and the vector field $\mathbf{F}$ must be continuous on the patch for the ordinary flux integral to apply.

The Planar Flux Formula

For an oriented parametrized surface, flux is computed by integrating the vector field dotted with the oriented area element. In an affine parametrization, differentiating with respect to the parameters gives $\mathbf{r}_u=\mathbf{a}$ and $\mathbf{r}_v=\mathbf{b}$. Thus the oriented area vector is the same at every point.

Theorem: Let $\mathbf{r}(u,v)=\mathbf{p}+u\mathbf{a}+v\mathbf{b}$ parametrize a planar patch over a suitable domain $D$, with $\mathbf{a}$ and $\mathbf{b}$ linearly independent. If $\mathbf{F}$ is continuous on the patch, the flux in the orientation determined by the order $u,v$ is $$ \iint_S \mathbf{F}\cdot d\mathbf{S} = \iint_D \mathbf{F}\bigl(\mathbf{p}+u\mathbf{a}+v\mathbf{b}\bigr)\cdot(\mathbf{a}\times\mathbf{b})\,du\,dv. $$

The magnitude $\lVert\mathbf{a}\times\mathbf{b}\rVert$ is the area scaling from parameter space to the plane. The unit normal in this orientation is $(\mathbf{a}\times\mathbf{b})/\lVert\mathbf{a}\times\mathbf{b}\rVert$, but the flux formula uses the full cross product, not just that unit vector.

Why the Cross Product Gives the Area Vector

Consider a small rectangle in parameter space with side lengths $\Delta u$ and $\Delta v$. Under the affine map, its sides become $\mathbf{a}\Delta u$ and $\mathbf{b}\Delta v$. The oriented area vector of the resulting parallelogram is

(\mathbf{a}\Delta u)\times(\mathbf{b}\Delta v) = (\mathbf{a}\times\mathbf{b})\Delta u\,\Delta v.

On this small piece, the vector field is approximately its value at a point $\mathbf{r}(u,v)$. The approximate flux is therefore $\mathbf{F}(\mathbf{r}(u,v))\cdot(\mathbf{a}\times\mathbf{b})\Delta u\,\Delta v$. Adding these contributions over a subdivision of $D$ and taking the limit gives the double integral in the theorem. Since both tangent vectors are constant, no position-dependent area vector appears in that limit.

This reasoning also shows why the parameter order specifies orientation. Reversing the order changes $\mathbf{a}\times\mathbf{b}$ to $\mathbf{b}\times\mathbf{a}=-(\mathbf{a}\times\mathbf{b})$, and hence reverses the flux sign while leaving the geometric patch unchanged.

Worked Examples

Worked Example: Flux Over a Parallelogram

Find the flux of $\mathbf{F}(x,y,z)=\langle x,y^2,z\rangle$ through the patch parametrized by $\mathbf{r}(u,v)=\langle 1,0,2\rangle+u\langle 2,1,0\rangle+v\langle 0,1,1\rangle$, for $0\leq u\leq1$, $0\leq v\leq1$, with the orientation given by $u,v$.

The fixed vectors are $\mathbf{a}=\langle2,1,0\rangle$ and $\mathbf{b}=\langle0,1,1\rangle$. Their cross product is

\mathbf{a}\times\mathbf{b} = \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 2&1&0\\ 0&1&1 \end{vmatrix} = \langle 1,-2,2\rangle.

The point on the patch is $(x,y,z)=(1+2u,u+v,2+v)$. Substituting this into the field gives $\mathbf{F}(\mathbf{r}(u,v))=\langle1+2u,(u+v)^2,2+v\rangle$. Its dot product with the oriented area vector is

\begin{aligned} \mathbf{F}(\mathbf{r}(u,v))\cdot(\mathbf{a}\times\mathbf{b}) &=(1+2u)-2(u+v)^2+2(2+v)\\ &=5+2u+2v-2u^2-4uv-2v^2. \end{aligned}

Integrate term by term over the unit square:

\begin{aligned} \text{Flux} &=\int_0^1\int_0^1 \left(5+2u+2v-2u^2-4uv-2v^2\right)\,dv\,du\\ &=5+1+1-\frac{2}{3}-1-\frac{2}{3}\\ &=\frac{14}{3}. \end{aligned}

For the check, the six terms integrate respectively to $5$, $1$, $1$, $-2/3$, $-1$, and $-2/3$; their sum is $14/3$. The positive result is the net flux in the orientation given by $u,v$.

Worked Example: Flux Over a Triangular Parameter Domain

Find the flux of $\mathbf{F}(x,y,z)=\langle x^2,y,z\rangle$ through $\mathbf{r}(u,v)=\langle u,v,u+2v\rangle$, where $D$ is the triangle $u\geq0,\ v\geq0,\ u+v\leq1$, with orientation $u,v$.

Here $\mathbf{a}=\langle1,0,1\rangle$ and $\mathbf{b}=\langle0,1,2\rangle$, so

\mathbf{a}\times\mathbf{b} = \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 1&0&1\\ 0&1&2 \end{vmatrix} = \langle-1,-2,1\rangle.

Since $x=u$, $y=v$, and $z=u+2v$, the field on the patch is $\langle u^2,v,u+2v\rangle$. Its dot product with the area vector simplifies to

\langle u^2,v,u+2v\rangle\cdot\langle-1,-2,1\rangle =-u^2-2v+u+2v =u-u^2.

Using $0\leq u\leq1$ and $0\leq v\leq1-u$, the flux is

\begin{aligned} \text{Flux} &=\int_0^1\int_0^{1-u}(u-u^2)\,dv\,du\\ &=\int_0^1(u-u^2)(1-u)\,du\\ &=\int_0^1(u-2u^2+u^3)\,du\\ &=\frac12-\frac23+\frac14 =\frac1{12}. \end{aligned}

As a check, the integral of $u-u^2$ across the triangle is positive: the integrand is nonnegative for $0\leq u\leq1$, and the computed value is $1/12$. In particular, the cancellation of the $v$-terms in the dot product is consistent with the result.

Worked Example: Carefully Expanding the Flux Integrand

Consider the upward-oriented unit square patch $\mathbf{r}(u,v)=\langle u,v,0\rangle$, with $0\leq u,v\leq1$. Let $\mathbf{F}(x,y,z)=\left\langle0,0,(-9-2x)(1-x)-\frac{19}{2}(1-x)^2\right\rangle$. Compute its flux.

The tangent vectors are $\mathbf{a}=\langle1,0,0\rangle$ and $\mathbf{b}=\langle0,1,0\rangle$, so $\mathbf{a}\times\mathbf{b}=\langle0,0,1\rangle$. On the patch, $x=u$, and the dot product is

(-9-2u)(1-u)-\frac{19}{2}(1-u)^2.

Expand each product before integrating:

\begin{aligned} (-9-2u)(1-u)&=-9+7u+2u^2,\\ -\frac{19}{2}(1-u)^2&=-\frac{19}{2}+19u-\frac{19}{2}u^2. \end{aligned}

Adding like terms gives $-37/2+26u-15u^2/2$. The integrand does not depend on $v$, whose interval has length one, so

\begin{aligned} \text{Flux} &=\int_0^1\left(-\frac{37}{2}+26u-\frac{15}{2}u^2\right)\,du\\ &=\left[-\frac{37}{2}u+13u^2-\frac{5}{2}u^3\right]_0^1\\ &=-\frac{37}{2}+13-\frac52\\ &=-8. \end{aligned}

Check the expansion at $u=1$: the original expression is zero, and the polynomial gives $-37/2+26-15/2=0$. The antiderivative also differentiates to $-37/2+26u-15u^2/2$, confirming the integral calculation.

Orientation and the Area Factor

Suppose a prescribed unit normal $\mathbf{n}$ is given. To check whether the parameter order agrees with it, calculate $(\mathbf{a}\times\mathbf{b})\cdot\mathbf{n}$. A positive value means the cross product points in the prescribed direction; a negative value means it points the opposite way. If the orientation is opposite, reverse the order of the parameter vectors or negate the oriented area vector.

A common mistake is to replace the cross product by a unit normal but leave out the area scaling. In the first example, the cross product has magnitude $\sqrt{1+4+4}=3$. Using only the unit normal $\langle1/3,-2/3,2/3\rangle$ in an integral over the unit parameter square would give $14/9$, not the correct $14/3$. The missing factor is $3$: the patch’s area scaling is $\lVert\mathbf{a}\times\mathbf{b}\rVert=3$. Using the full vector $\mathbf{a}\times\mathbf{b}$ includes that factor automatically.

Takeaway: For an affine planar patch, first compute $\mathbf{a}\times\mathbf{b}$, then substitute the parametrization into the field, take the dot product, and integrate over $D$. Keep the parameter order consistent with the requested orientation.

When This Method Is Useful

This formula is especially efficient when a planar patch is naturally described as a parallelogram or as the image of a simple parameter region. The plane’s normal direction and area scaling are fixed, so the main remaining work is evaluating the field on the patch and integrating the resulting scalar function over $D$. Rectangles give iterated integrals with constant bounds; triangles often use bounds such as $0\leq u\leq1$ and $0\leq v\leq1-u$.

The method also provides a practical check on sign and scale. A negative flux means the field’s average normal component points against the selected orientation; it does not imply that the area is negative. If the cross product is reversed, the flux changes sign, while the unsigned area of the patch remains unchanged.

Check Your Understanding

For each question, identify the constant oriented area vector before setting up the integral.

For $\mathbf{r}(u,v)=\langle 2,1,0\rangle+u\langle1,2,0\rangle+v\langle0,0,3\rangle$, what is the oriented area vector for parameter order $u,v$?
Why does an affine parametrization of a plane patch have a constant oriented area vector?
What condition on $\mathbf{a}$ and $\mathbf{b}$ ensures the affine parametrization describes a two-dimensional patch?
What changes in the flux formula if the parameter order is reversed?
Why is it generally incorrect to use a unit normal in place of $\mathbf{a}\times\mathbf{b}$ without another area factor?

What You'll Learn