Stokes' Theorem on Graph Surfaces

From a Graph Surface to Its Projection

A surface given by \(z=g(x,y)\) is often easier to work with through its projection onto the \(xy\)-plane. Stokes' theorem makes this useful: instead of integrating curl over the curved surface, we can express the flux integral as a double integral over the planar region beneath it.

The key is to get the orientation right. The graph determines a normal direction, and that normal direction determines how its boundary must be traversed. A sign error in the normal element changes the sign of the flux and corresponds to reversing the boundary orientation.

Definition: Let \(D\) be a region in the \(xy\)-plane and let \(g(x,y)\) be a differentiable function on \(D\). The graph surface over \(D\) is \(S=\{(x,y,g(x,y)):(x,y)\in D\}\). Its projection onto the \(xy\)-plane is the region \(D\).

Deriving the Upward Normal Element

Parametrize the graph using its planar coordinates:

\mathbf r(x,y)=\langle x,y,g(x,y)\rangle,\qquad (x,y)\in D.

The two coordinate tangent vectors are obtained by differentiating with respect to \(x\) and \(y\):

\mathbf r_x=\langle 1,0,g_x\rangle, \qquad \mathbf r_y=\langle 0,1,g_y\rangle.

Their cross product is a normal vector to the surface. Choosing the order \(\mathbf r_x\times\mathbf r_y\) gives a normal with positive \(z\)-component:

\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} = \langle -g_x,-g_y,1\rangle.

The positive third component means this normal points upward. The vector surface element, which combines the normal direction and the area scaling, is therefore \((\mathbf r_x\times\mathbf r_y)\,dx\,dy\).

Key formula: For the upward orientation of the graph \(z=g(x,y)\), the vector surface element is \(\mathbf n\,dS=\langle -g_x,-g_y,1\rangle\,dx\,dy\).

This is a vector area element, not just a unit normal. Its length accounts for the stretching of the graph relative to its projection. In fact, its magnitude is \(\sqrt{1+g_x^2+g_y^2}\), so the scalar area element is \(dS=\sqrt{1+g_x^2+g_y^2}\,dx\,dy\), while the vector element retains the direction needed for flux.

Stokes' Theorem on a Graph

Suppose \(D\) is bounded with piecewise-smooth boundary, \(g\) is twice continuously differentiable on a neighborhood of \(D\), and the vector field \(\mathbf F\) has continuous first partial derivatives on a neighborhood of the graph surface. Stokes' theorem states that the circulation of \(\mathbf F\) around the oriented boundary equals the flux of \(\nabla\times\mathbf F\) through the oriented surface.

Theorem: For the upward-oriented graph \(S\) over \(D\), let \(C=\partial S\) have the induced boundary orientation. Then \(\displaystyle \oint_C\mathbf F\cdot d\mathbf r=\iint_S(\nabla\times\mathbf F)\cdot\mathbf n\,dS =\iint_D(\nabla\times\mathbf F)(x,y,g(x,y))\cdot\langle-g_x,-g_y,1\rangle\,dx\,dy.\)

In the last integral, the curl is evaluated at points on the graph, so wherever \(\mathbf F\) depends on \(z\), substitute \(z=g(x,y)\). The theorem does not say that the planar region itself is the surface; it provides a way to calculate the flux through the graph using coordinates in its projection.

The boundary direction must match the upward normal. For a region with one boundary component and no holes, the projected boundary is traversed counterclockwise as viewed from above. If \(D\) has holes, its outer boundary components are counterclockwise and the boundaries of the holes are clockwise. This gives the orientation induced by one upward normal across the graph.

A Reliable Setup

Before integrating, make the geometry and sign choices explicit. The following sequence keeps the surface and boundary orientations consistent.

Identify the projected region.
Write \(D\) in the \(xy\)-plane and record its boundary components.

Find the graph derivatives.
Compute \(g_x\) and \(g_y\) for \(z=g(x,y)\).

Choose the upward normal element.
Use \(\langle-g_x,-g_y,1\rangle\,dx\,dy\), whose third component is positive.

Evaluate the curl on the graph.
Compute \(\nabla\times\mathbf F\), then substitute \(z=g(x,y)\) in its components.

Set up the projected integral.
Take the dot product with the upward normal element and integrate over \(D\).

Worked Examples

Worked Example: A Paraboloid Over a Disk

Let \(S\) be the graph \(z=x^2+y^2\) over the unit disk \(D=\{(x,y):x^2+y^2\leq1\}\), oriented upward. Find the curl flux for \(\mathbf F=\langle-y/2,x/2,0\rangle\), and check it using the boundary circulation.

First, \(g_x=2x\) and \(g_y=2y\), so the upward vector surface element is \(\langle-2x,-2y,1\rangle\,dx\,dy\). The curl is

\nabla\times\mathbf F = \left\langle \frac{\partial 0}{\partial y}-\frac{\partial (x/2)}{\partial z}, \frac{\partial (-y/2)}{\partial z}-\frac{\partial 0}{\partial x}, \frac{\partial (x/2)}{\partial x}-\frac{\partial (-y/2)}{\partial y} \right\rangle = \langle0,0,1\rangle.

Its dot product with the upward normal element is \(1\). Therefore, using polar coordinates on the unit disk,

\iint_S(\nabla\times\mathbf F)\cdot\mathbf n\,dS = \iint_D1\,dx\,dy = \int_0^{2\pi}\int_0^1 r\,dr\,d\theta = \int_0^{2\pi}\left[\frac{r^2}{2}\right]_0^1d\theta = \int_0^{2\pi}\frac12\,d\theta = \pi.

The boundary projects counterclockwise, so parametrize the lifted boundary by \(\mathbf r(t)=\langle\cos t,\sin t,1\rangle\), \(0\leq t\leq2\pi\). Then \(d\mathbf r=\langle-\sin t,\cos t,0\rangle\,dt\) and \(\mathbf F(\mathbf r(t))=\langle-\sin t/2,\cos t/2,0\rangle\). Thus

\oint_C\mathbf F\cdot d\mathbf r = \int_0^{2\pi} \left(\frac{\sin^2t}{2}+\frac{\cos^2t}{2}\right)dt = \int_0^{2\pi}\frac12\,dt = \pi.

The boundary circulation equals the curl flux, as Stokes' theorem predicts.

Worked Example: A Bilinear Graph Over a Rectangle

Let \(z=xy\) over \(D=[0,1]\times[0,2]\), with upward orientation, and take \(\mathbf F=\langle0,0,x\rangle\). Set up and evaluate the curl flux, then verify it around the boundary.

Here \(g_x=y\) and \(g_y=x\), so the upward normal element is \(\langle-y,-x,1\rangle\,dx\,dy\). Directly from the component formula for curl,

\nabla\times\mathbf F = \left\langle \frac{\partial x}{\partial y}-\frac{\partial 0}{\partial z}, \frac{\partial 0}{\partial z}-\frac{\partial x}{\partial x}, \frac{\partial 0}{\partial x}-\frac{\partial 0}{\partial y} \right\rangle = \langle0,-1,0\rangle.

The dot product with the upward normal is \(x\). Therefore,

\iint_D x\,dx\,dy = \int_0^2\int_0^1 x\,dx\,dy = \int_0^2\left[\frac{x^2}{2}\right]_0^1dy = \int_0^2\frac12\,dy = 1.

For the circulation check, traverse the rectangle counterclockwise in projection. On the bottom edge, \(y=0\), \(z=0\), so \(dz=0\) and the contribution \(\int x\,dz\) is \(0\). On the right edge, \(x=1\), \(y\) runs from \(0\) to \(2\), and \(z=y\); its contribution is \(\int_0^2 1\,dy=2\). On the top edge, \(y=2\), \(x\) runs from \(1\) to \(0\), and \(z=2x\); its contribution is \(\int_1^0 2x\,dx=-1\). On the left edge, \(x=0\), so the contribution is \(0\). Hence the total circulation is \(0+2-1+0=1\), agreeing with the flux.

Worked Example: A Triangular Projection

Let \(S\) be the graph \(z=x+y\) over the triangle \(D=\{(x,y):0\leq x\leq1,\ 0\leq y\leq1-x\}\), oriented upward. Use \(\mathbf F=\langle0,0,x^2\rangle\) to find the curl flux and check the result around the lifted boundary.

The derivatives are \(g_x=1\) and \(g_y=1\), so the normal element is \(\langle-1,-1,1\rangle\,dx\,dy\). The curl is

\nabla\times\mathbf F = \left\langle \frac{\partial (x^2)}{\partial y}-\frac{\partial 0}{\partial z}, \frac{\partial 0}{\partial z}-\frac{\partial (x^2)}{\partial x}, \frac{\partial 0}{\partial x}-\frac{\partial 0}{\partial y} \right\rangle = \langle0,-2x,0\rangle.

The dot product with \(\langle-1,-1,1\rangle\) is \(2x\). Integrating over the triangle gives

\iint_D2x\,dx\,dy = \int_0^1\int_0^{1-x}2x\,dy\,dx = \int_0^1 2x(1-x)\,dx = 2\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1 = 2\left(\frac12-\frac13\right) = \frac13.

The projected boundary runs from \((0,0)\) to \((1,0)\), then to \((0,1)\), then back to \((0,0)\). On the first edge, \(y=0\), \(x\) goes from \(0\) to \(1\), and \(z=x\). Since \(F_z=x^2\), its contribution is \(\int_0^1x^2\,dx=1/3\). On the diagonal edge, \(x=1-t,\ y=t\) for \(0\leq t\leq1\); then \(z=x+y=1\), so \(dz=0\) and the contribution is \(0\). On the final edge, \(x=0\), so \(F_z=x^2=0\) and the contribution is \(0\). The circulation is \(1/3+0+0=1/3\), matching the flux.

Orientation and Sign Pitfalls

A common error is to reverse the cross product without changing the orientation. For the graph \(z=x+y\), the correct upward normal element is \(\langle-1,-1,1\rangle\,dx\,dy\). Reversing the cross product gives \(\langle1,1,-1\rangle\,dx\,dy\), which points downward. In the triangular example, that reversed vector would give the integrand \(-2x\) and the flux \(-1/3\). The calculation is not consistent with the upward orientation and the counterclockwise projected boundary; it describes the opposite orientation.

Another mistake is to say that every boundary component projects counterclockwise. That is correct for the outer boundary of a region without holes, but not for the boundary of a hole. With an upward normal, a hole boundary is traversed clockwise as viewed from above. For example, an annular projection has a counterclockwise outer circle and a clockwise inner circle. Reversing the inner direction would not be the boundary orientation induced by the same upward normal.

Orientation check: The sign of the third component in \(\langle-g_x,-g_y,1\rangle\) confirms that the normal points upward. The projected boundary directions must be consistent with that orientation: outer components counterclockwise and hole boundaries clockwise.

Why the Projected Integral Is Useful

The graph formula reduces a surface flux calculation to an integral over a planar region, where familiar descriptions such as rectangles, disks, and triangles can make the bounds straightforward. Its advantage is not that the surface disappears: the graph still matters through \(g_x\), \(g_y\), and the substitution \(z=g(x,y)\) in the curl. It gives a direct way to organize those contributions while keeping the orientation visible.

When using the formula, check three items before evaluating: the graph derivatives in the normal element, the values of the curl on the graph, and the sign implied by the chosen orientation. A useful final check, when the boundary integral is manageable, is to compute the circulation using the induced boundary direction and compare the two results.

Check Your Understanding

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

For \(z=g(x,y)\), what is the upward-oriented vector surface element, and why does it point upward?
What substitution is required when a component of \(\nabla\times\mathbf F\) depends on \(z\)?
If the projected region is an annulus, how are its outer and inner boundary components oriented for the upward normal?
What orientation results from using \(\mathbf r_y\times\mathbf r_x\) instead of \(\mathbf r_x\times\mathbf r_y\)?
For the graph \(z=x+y\) over the triangle in the third example, explain why the diagonal boundary contributes zero to the circulation of \(\langle0,0,x^2\rangle\).