Stokes Along Curves of Intersecting Surfaces

Introduction

A closed curve need not be given by a familiar circle or ellipse in a coordinate plane. It may instead be specified as the intersection of two surfaces, such as a plane and a cylinder. The curve can look spatially complicated, but its geometry often supplies a particularly simple surface that it bounds.

For a vertical circular cylinder and a nonvertical plane, the intersection is a closed curve whose horizontal projection is a circle. The curve itself lies in the plane, so it bounds a planar disk there. That disk gives a direct route to the circulation: orient it, compute the curl of the vector field, and integrate the curl’s normal component over the disk.

Parameterizing the Intersection

Consider the circular cylinder \(x^2+y^2=R^2\), where \(R>0\), and the plane \(z=ax+by+c\), where \(a\), \(b\), and \(c\) are constants. On the cylinder, the horizontal coordinates can be written as \(x=R\cos t\) and \(y=R\sin t\). Substitution into the plane equation gives the following parameterization of their intersection:

\mathbf{r}(t) = \bigl(R\cos t,\ R\sin t,\ aR\cos t+bR\sin t+c\bigr), \qquad 0\leq t\leq 2\pi.

As \(t\) increases from \(0\) to \(2\pi\), the projected point \((x,y)\) travels counterclockwise around the circle \(x^2+y^2=R^2\). The spatial curve closes because the parameterization has the same position at \(t=0\) and \(t=2\pi\). It is smooth and regular: its derivative has horizontal components \(-R\sin t\) and \(R\cos t\), which cannot both be zero when \(R>0\).

Definition: For \(R>0\), the intersection of \(x^2+y^2=R^2\) and \(z=ax+by+c\) is parameterized by \(\mathbf{r}(t)\) above. Increasing \(t\) gives the direction whose projection onto the \(xy\)-plane is counterclockwise.

The Planar Disk and Its Orientation

The intersection lies in the plane \(z=ax+by+c\). The part of that plane inside the cylinder is a disk, parameterized over the ordinary disk \(D=\{(x,y):x^2+y^2\leq R^2\}\) by \(\mathbf{q}(x,y)=(x,y,ax+by+c)\). Its tangent vectors in the coordinate directions are \(\mathbf{q}_x=(1,0,a)\) and \(\mathbf{q}_y=(0,1,b)\). Their cross product is \((-a,-b,1)\), which points upward because its \(z\)-component is positive.

Thus the oriented area element for this upward orientation is \((-a,-b,1)\,dx\,dy\). Its boundary direction is the one induced by the right-hand rule: when viewed from above, the boundary goes counterclockwise. That is exactly the direction obtained by increasing \(t\) in \(\mathbf{r}(t)\).

\mathbf{q}_x\times\mathbf{q}_y=(-a,-b,1), \qquad d\mathbf{S}=(-a,-b,1)\,dx\,dy.

Stokes’ Theorem states that, for an oriented smooth surface \(S\) with positively oriented boundary curve \(C\), and a vector field \(\mathbf{F}\) with continuous first partial derivatives on a neighborhood of \(S\),

\oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_S (\nabla\times\mathbf{F})\cdot d\mathbf{S}.

Here, \(S\) is the planar disk and \(C\) is its circular-cylinder intersection. With the upward normal, the increasing-\(t\) direction is positive. The theorem therefore converts the circulation around the spatial curve into a double integral over the disk. This setup requires the field to be defined and continuously differentiable on a neighborhood of the disk; if the field has a singularity there, the theorem cannot be applied in this way.

Why the Area Vector Gives the Right Flux

The cross product \(\mathbf{q}_x\times\mathbf{q}_y\) accounts for both the plane’s tilt and the surface’s orientation. A small rectangle of area \(dx\,dy\) in the \(xy\)-plane maps to a small parallelogram on the tilted plane. The cross product of its two edge directions gives the parallelogram’s oriented area vector per unit \(xy\)-area. Consequently, dotting the curl with \((-a,-b,1)\) gives the curl component normal to the plane, with the correct area scaling already included.

In particular, there is no need to calculate the planar disk’s actual area in the plane and then separately adjust for its tilt. The vector area element already performs that adjustment. Applying Stokes’ Theorem in these coordinates gives

\oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_D (\nabla\times\mathbf{F})(x,y,ax+by+c) \cdot(-a,-b,1)\,dx\,dy.

This identity follows by parameterizing the disk with \(\mathbf{q}\), using \(\mathbf{q}_x\times\mathbf{q}_y=(-a,-b,1)\) in the surface integral, and using the counterclockwise boundary direction already matched to increasing \(t\).

Worked Examples

Worked Example: A Tilted Plane and a Constant Curl

Let \(C\) be the intersection of \(x^2+y^2=4\) and \(z=x+2y+1\), directed by increasing \(t\). Find the circulation of \(\mathbf{F}(x,y,z)=(-y/2,x/2,0)\).

The plane has \(a=1\), \(b=2\), \(c=1\), and the cylinder has \(R=2\). Its upward-oriented disk has area vector \((-1,-2,1)\,dx\,dy\). First compute the curl:

\nabla\times\mathbf{F} = \left( \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) =(0,0,1).

The curl’s dot product with the oriented area vector is \((0,0,1)\cdot(-1,-2,1)=1\). The projected disk has radius \(2\), so its area is \(\pi(2)^2=4\pi\). Therefore Stokes’ Theorem gives

\oint_C\mathbf{F}\cdot d\mathbf{r} = \iint_D 1\,dx\,dy = 4\pi.

Check by direct parameterization. Here \(\mathbf{r}(t)=(2\cos t,2\sin t,2\cos t+4\sin t+1)\), and \(\mathbf{r}'(t)=(-2\sin t,2\cos t,-2\sin t+4\cos t)\). Along the curve, \(\mathbf{F}(\mathbf{r}(t))=(-\sin t,\cos t,0)\). Hence

\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t) = 2\sin^2 t+2\cos^2 t=2, \qquad \int_0^{2\pi}2\,dt=4\pi.

The direct line integral agrees with the Stokes calculation.

Worked Example: A Nonconstant Curl Over a Unit Disk

Let \(C\) be the intersection of \(x^2+y^2=1\) and \(z=2x-y\), directed by increasing \(t\). Compute the circulation of \(\mathbf{F}(x,y,z)=(-y(x^2+y^2)/2,\ x(x^2+y^2)/2,\ 0)\).

For this plane, \(a=2\) and \(b=-1\), so the upward area vector is \((-2,1,1)\,dx\,dy\). Write \(P=-y(x^2+y^2)/2\) and \(Q=x(x^2+y^2)/2\). Since the field’s third component is zero and the first two components do not depend on \(z\), the first two curl components are zero. The third is

\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} = \frac{3x^2+y^2}{2} - \left(-\frac{x^2+3y^2}{2}\right) = 2(x^2+y^2).

Thus \(\nabla\times\mathbf{F}=(0,0,2(x^2+y^2))\). Its dot product with \((-2,1,1)\) is \(2(x^2+y^2)\). In polar coordinates on the unit disk, \(x^2+y^2=r^2\) and \(dx\,dy=r\,dr\,d\theta\). The circulation is therefore

\iint_D 2(x^2+y^2)\,dx\,dy = \int_0^{2\pi}\int_0^1 2r^2\,r\,dr\,d\theta = \int_0^{2\pi}\left[\frac{r^4}{2}\right]_0^1d\theta = \int_0^{2\pi}\frac12\,d\theta = \pi.

To check directly, parameterize \(C\) by \(\mathbf{r}(t)=(\cos t,\sin t,2\cos t-\sin t)\). On the curve \(x^2+y^2=1\), so \(\mathbf{F}(\mathbf{r}(t))=(-\sin t/2,\cos t/2,0)\), and \(\mathbf{r}'(t)=(-\sin t,\cos t,-2\sin t-\cos t)\). Their dot product is \((\sin^2 t+\cos^2 t)/2=1/2\). Its integral from \(0\) to \(2\pi\) is \(\pi\), as required.

Worked Example: Reversing the Curve Direction

Let \(C\) be the intersection of \(x^2+y^2=1\) and \(z=x+2y\). First calculate the circulation of \(\mathbf{F}(x,y,z)=(0,0,x)\) in the direction given by increasing \(t\), then find it in the opposite direction.

The upward area vector is \((-1,-2,1)\,dx\,dy\). The curl is

\nabla\times\mathbf{F} = \left( \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) =(0,-1,0).

The curl dotted with the area vector is \((0,-1,0)\cdot(-1,-2,1)=2\). The projected disk is the unit disk, with area \(\pi\). Thus the increasing-\(t\) circulation is \(2\pi\). Reversing the direction of a line integral changes its sign, so the opposite direction gives \(-2\pi\).

For a direct check, \(\mathbf{r}(t)=(\cos t,\sin t,\cos t+2\sin t)\), so \(z'(t)=-\sin t+2\cos t\). Since \(\mathbf{F}\cdot d\mathbf{r}=x\,dz\),

\int_0^{2\pi}\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt = \int_0^{2\pi}\cos t(-\sin t+2\cos t)\,dt = 0+2\pi = 2\pi.

The first term integrates to zero because \(\int_0^{2\pi}\sin t\cos t\,dt=0\); the second integrates to \(2\int_0^{2\pi}\cos^2t\,dt=2\pi\). Reversing the limits, or equivalently reversing the parameter direction, yields \(-2\pi\), confirming the orientation sign.

A Common Orientation Mistake

The direction of the intersection curve cannot be chosen independently of the surface normal in Stokes’ Theorem. For the upward normal \((-a,-b,1)\), the positive boundary direction projects counterclockwise onto the \(xy\)-plane. If a calculation instead uses the decreasing-\(t\) direction, its circulation has the opposite sign.

For instance, in the last example the upward-oriented disk gives the value \(2\pi\). A mistaken sign calculation might use the upward normal but report \(-2\pi\) for the increasing-\(t\) curve. The correction is to check the projected direction: increasing \(t\) moves counterclockwise, which matches the upward normal, so the Stokes value is \(2\pi\). The value \(-2\pi\) belongs to the reversed curve, not to the stated orientation.

Orientation check: For \(z=ax+by+c\) with upward normal \((-a,-b,1)\), increasing \(t\) in \(\mathbf{r}(t)\) is the positive boundary direction. Reversing the curve direction negates the circulation.

Why This Setup Is Useful

A curve described as the intersection of surfaces may be awkward to use in a direct line integral: each component of \(\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\) can contain several trigonometric terms. The plane containing the curve supplies a disk with a simple projected domain, and Stokes’ Theorem replaces that line integral with a curl integral over the disk. The conversion is especially effective when the curl is constant, when its normal component is simple, or when symmetry makes parts of the disk integral vanish.

The geometric steps are consistent: parameterize the intersection, identify the bounded planar patch, compute its oriented area vector, and make sure the curve direction matches that normal. Once those choices agree, the theorem gives the circulation without requiring a direct integration along the spatial curve.

Check Your Understanding

Use the curve and orientation information in each question to identify the appropriate Stokes setup before calculating.

For \(x^2+y^2=9\) and \(z=2x+3y-1\), write a parameterization of the intersection in the direction whose projection is counterclockwise.
What upward-oriented area vector belongs to the plane \(z=-x+4y+2\), when the surface is parameterized over the \(xy\)-plane?
For \(z=3x-y\), does increasing \(t\) in the standard cylinder parameterization agree with the upward normal’s boundary orientation?
If the curve direction in a Stokes circulation is reversed while the surface and field remain fixed, how does the answer change?
For a field with continuous first partial derivatives near the planar disk, which integral does Stokes’ Theorem equate to the circulation around the cylinder–plane intersection?