Flux Through Graphs Over the YZ-Plane

Graphs Over the \(yz\)-Plane

A surface can be represented as a graph over the \(yz\)-plane when its \(x\)-coordinate is determined by the other two coordinates. In this setting, \(y\) and \(z\) vary over a region \(D\), and the surface has equation \(x=g(y,z)\). The flux calculation follows the same central principle as for other parametrized surfaces: integrate the vector field's component in the oriented normal direction, including the correct area scaling.

The important orientation choice here is whether the normal points toward positive or negative \(x\). Deriving the area vector directly makes this choice explicit and prevents an unnoticed sign change.

Definition: Let \(D\) be a region in the \(yz\)-plane and let \(g:D\to\mathbb{R}\) have continuous first partial derivatives. The graph surface is \(S=\{(g(y,z),y,z):(y,z)\in D\}\). Its positive \(x\)-orientation is the orientation whose normal vector has positive \(x\)-component; its negative \(x\)-orientation has negative \(x\)-component.

Deriving the Oriented Area Vector

Use \(y\) and \(z\), in that order, as parameters. A parametrization is \(\mathbf{r}(y,z)=(g(y,z),y,z)\). Its tangent vectors point in the directions in which the surface changes as one parameter varies while the other is held fixed. Taking their cross product gives an area vector for a small parameter rectangle.

\mathbf{r}_y=(g_y,1,0), \qquad \mathbf{r}_z=(g_z,0,1).

With the parameter order \((y,z)\), the cross product \(\mathbf{r}_y\times\mathbf{r}_z\) is the oriented area vector per unit area in the parameter domain. Computing its components gives

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

This vector points in the positive \(x\)-direction because its \(x\)-component is \(1\). Its magnitude, \(\sqrt{1+g_y^2+g_z^2}\), accounts for the surface area associated with a unit area in the \(yz\)-plane. Reversing the orientation negates the whole vector, not just its \(x\)-component.

Result: For the parametrization \(\mathbf{r}(y,z)=(g(y,z),y,z)\), the oriented area vectors are \((1,-g_y,-g_z)\,dy\,dz\) for positive \(x\)-orientation and \((-1,g_y,g_z)\,dy\,dz\) for negative \(x\)-orientation.

The Flux Formulas

Write a vector field as \(\mathbf{F}(x,y,z)=(P(x,y,z),Q(x,y,z),R(x,y,z))\). Flux through an oriented surface is the integral of the field dotted with its oriented area vector. Since points on the graph have \(x=g(y,z)\), evaluate all three components of the field there.

\begin{aligned} \text{Positive }x\text{-orientation:}\quad \iint_S \mathbf{F}\cdot d\mathbf{S} &= \iint_D \mathbf{F}(g(y,z),y,z)\cdot(1,-g_y,-g_z)\,dy\,dz\\ &= \iint_D \bigl(P(g(y,z),y,z)-Q(g(y,z),y,z)g_y -R(g(y,z),y,z)g_z\bigr)\,dy\,dz. \end{aligned}

\text{Negative }x\text{-orientation:}\qquad \iint_S \mathbf{F}\cdot d\mathbf{S} = -\iint_D \bigl(P(g(y,z),y,z)-Q(g(y,z),y,z)g_y -R(g(y,z),y,z)g_z\bigr)\,dy\,dz.

These formulas require a differentiable graph, a specified parameter region \(D\), and a vector field defined and continuous on the surface so the flux integral exists. The domain is integrated with respect to \(dy\,dz\), since its coordinates are \(y\) and \(z\). The factor \(\sqrt{1+g_y^2+g_z^2}\) does not need to be added separately: the unnormalized area vector already includes the area scaling.

Why the Formula Works

The cross product both measures the area of a small surface patch and records its orientation. To see why it gives the flux element, consider a small rectangle in the parameter domain with side lengths \(\Delta y\) and \(\Delta z\). Its image on the surface is approximately a parallelogram with side vectors \(\mathbf{r}_y\Delta y\) and \(\mathbf{r}_z\Delta z\). The parallelogram's oriented area vector is their cross product.

(\mathbf{r}_y\Delta y)\times(\mathbf{r}_z\Delta z) = (\mathbf{r}_y\times\mathbf{r}_z)\Delta y\Delta z.

Over such a small patch, the field is approximately constant, so the flux is approximately the field dotted with this area vector. Adding these contributions over the domain and taking the limit gives the surface integral. For the graph, the cross product is \((1,-g_y,-g_z)\), so it points toward positive \(x\); negating it produces the other orientation. This establishes both the area factor and the sign in the formulas.

Worked Examples

Worked Example: Positive \(x\)-Flux Over a Rectangle

Find the flux of \(\mathbf{F}(x,y,z)=(x^2,y,z)\) through \(x=y+z\), over \(0\le y\le1\), \(0\le z\le1\), with positive \(x\)-orientation.

Here \(g(y,z)=y+z\), so \(g_y=1\) and \(g_z=1\). The positive \(x\)-oriented area vector is \((1,-1,-1)\). On the graph, the field is \(\mathbf{F}(y+z,y,z)=((y+z)^2,y,z)\). Thus the flux integrand is

((y+z)^2,y,z)\cdot(1,-1,-1) =(y+z)^2-y-z =y^2+2yz+z^2-y-z.

Integrate first with respect to \(z\):

\begin{aligned} \int_0^1 (y^2+2yz+z^2-y-z)\,dz &=y^2+y+\frac13-y-\frac12\\ &=y^2-\frac16. \end{aligned}

Then integrate with respect to \(y\):

\int_0^1\left(y^2-\frac16\right)\,dy =\frac13-\frac16 =\frac16.

The flux is \(\frac16\). As a sign check, reversing the area vector to \((-1,1,1)\) changes the dot product to the negative of the integrand above, so the reversed flux would be \(-\frac16\), as required for the opposite orientation.

Worked Example: Flux Over a Circular Domain

Find the flux of \(\mathbf{F}(x,y,z)=(x,y,z)\) through \(x=y^2+z^2\), over the unit disk \(D=\{(y,z):y^2+z^2\le1\}\), with negative \(x\)-orientation.

Here \(g_y=2y\) and \(g_z=2z\). The positive \(x\)-oriented area vector would be \((1,-2y,-2z)\), so the negative \(x\)-oriented one is \((-1,2y,2z)\). On the graph, \(\mathbf{F}=(y^2+z^2,y,z)\). Its dot product with the negative area vector is

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

Using polar coordinates in the \(yz\)-plane, set \(y=r\cos\theta\), \(z=r\sin\theta\); then \(y^2+z^2=r^2\), and \(dy\,dz=r\,dr\,d\theta\). The disk has \(0\le r\le1\) and \(0\le\theta\le2\pi\). Therefore,

\begin{aligned} \iint_D (y^2+z^2)\,dy\,dz &=\int_0^{2\pi}\int_0^1 r^2r\,dr\,d\theta\\ &=\int_0^{2\pi}\left[\frac{r^4}{4}\right]_0^1\,d\theta\\ &=\int_0^{2\pi}\frac14\,d\theta =\frac{\pi}{2}. \end{aligned}

The negative \(x\)-oriented flux is \(\frac{\pi}{2}\). As a check on the sign, the positive \(x\)-oriented area vector is its negative, so that orientation gives flux \(-\frac{\pi}{2}\). The two answers have equal magnitude and opposite signs.

Worked Example: A Mixed-Component Field

Find the positive \(x\)-oriented flux of \(\mathbf{F}(x,y,z)=(z,y,x)\) through \(x=yz\), over the rectangle \(0\le y\le1\), \(0\le z\le1\).

For \(g(y,z)=yz\), the partial derivatives are \(g_y=z\) and \(g_z=y\). The positive \(x\)-oriented area vector is therefore \((1,-z,-y)\). Substituting \(x=yz\) into the field gives \(\mathbf{F}(yz,y,z)=(z,y,yz)\). The integrand is

(z,y,yz)\cdot(1,-z,-y) =z-yz-y^2z =z(1-y-y^2).

Integrate over the rectangle, first with respect to \(z\):

\begin{aligned} \int_0^1 z(1-y-y^2)\,dz &=(1-y-y^2)\left[\frac{z^2}{2}\right]_0^1\\ &=\frac12(1-y-y^2). \end{aligned}

Now integrate with respect to \(y\):

\begin{aligned} \int_0^1\frac12(1-y-y^2)\,dy &=\frac12\left[y-\frac{y^2}{2}-\frac{y^3}{3}\right]_0^1\\ &=\frac12\left(1-\frac12-\frac13\right) =\frac1{12}. \end{aligned}

The flux is \(\frac1{12}\). Reversing the orientation changes the area vector from \((1,-z,-y)\) to \((-1,z,y)\), negating the dot product at every point; the flux for that orientation is consequently \(-\frac1{12}\).

A Common Orientation Error

A common mistake is to compute the cross product in the wrong order and then treat it as if it pointed toward positive \(x\). In the first example, the correct order gives

\mathbf{r}_y\times\mathbf{r}_z=(1,-1,-1).

But reversing the order gives

\mathbf{r}_z\times\mathbf{r}_y=(-1,1,1).

If this reversed vector is incorrectly labelled positive \(x\)-oriented, the calculation produces \(-\frac16\) instead of \(\frac16\). The correction is to check the \(x\)-component before using the vector: positive \(x\)-orientation requires a positive \(x\)-component. For the chosen parameter order, \(\mathbf{r}_y\times\mathbf{r}_z\) has that property; the reversed cross product represents negative \(x\)-orientation.

This check is especially useful because a cross product may have the right magnitude and still point in the wrong direction. Orientation is part of the flux calculation, not a separate cosmetic choice.

Check Your Understanding

Use the graph parametrization and orientation rules to answer the following questions.

For \(x=g(y,z)\), what is the cross product \(\mathbf{r}_y\times\mathbf{r}_z\) when \(\mathbf{r}(y,z)=(g(y,z),y,z)\)?
How does the area vector change when the orientation is reversed?
For positive \(x\)-orientation, what is the flux integrand in terms of \(P,Q,R,g_y,\) and \(g_z\), with the field evaluated on the graph?
Why should a separate factor \(\sqrt{1+g_y^2+g_z^2}\) not be inserted into the flux formula using the unnormalized area vector?
If a computed normal vector has negative \(x\)-component, what orientation does it represent?