Stokes' Theorem for Elliptical Boundaries

From an Elliptical Boundary to Its Interior

When a vector field is given along an ellipse, its circulation can be difficult to compute directly: substituting an elliptical parameterization into the line integral may produce a complicated expression. Stokes’ Theorem offers another route. It converts circulation around the boundary into the flux of the curl through a surface spanning that boundary.

For an ellipse in a horizontal plane, the simplest spanning surface is its flat interior. A useful change of variables then turns the elliptical region into a unit disk. The key is to scale the two circular coordinates by the ellipse’s two semiaxis lengths. This keeps the bounds simple while preserving the correct area through a Jacobian factor.

Central idea: For a positively oriented elliptical boundary, evaluate the $z$-component of the curl over the planar elliptical interior. Use coordinates scaled separately by the ellipse’s horizontal and vertical semiaxes.

Stokes’ Theorem and Ellipse Orientation

Write the vector field as $\mathbf{F}=(P,Q,R)$, where $P$, $Q$, and $R$ are functions of position with continuous first derivatives on a region containing the surface. Its curl has vertical component

(\nabla\times\mathbf{F})_z = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}.

Consider an ellipse in the plane $z=z_0$, centered at $(h,k,z_0)$, with semiaxis lengths $a>0$ and $b>0$. Its interior is described by

\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}\leq 1, \qquad z=z_0.

Choose the upward unit normal $\mathbf{n}=(0,0,1)$. The boundary orientation compatible with this normal is counterclockwise when viewed from above. Stokes’ Theorem says that the circulation around this oriented boundary equals the curl flux through the planar interior:

\oint_{\partial S}\mathbf{F}\cdot d\mathbf{r} = \iint_S(\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS.

Because the normal points in the $z$-direction, only the vertical component of the curl contributes. If the boundary is traversed clockwise instead, the orientation is reversed, and the circulation has the opposite sign.

Conditions to check: The field must have continuous first partial derivatives on a region containing the planar patch, including its boundary. The boundary direction and surface normal must be compatible: counterclockwise from above goes with the upward normal.

Scaled Circular Coordinates

To cover the full elliptical interior, use $0\leq r\leq 1$ and $0\leq\theta\leq 2\pi$, with

x=h+ar\cos\theta, \qquad y=k+br\sin\theta, \qquad z=z_0.

At $r=1$, these equations trace the ellipse’s boundary once counterclockwise. Values $0\leq r<1$ fill its interior. The two scale factors matter: $a$ stretches the circular coordinates in the $x$-direction, and $b$ stretches them in the $y$-direction.

The area change is determined by the Jacobian determinant of the transformation from $(r,\theta)$ to $(x,y)$:

\frac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix} a\cos\theta & -ar\sin\theta\\ b\sin\theta & br\cos\theta \end{vmatrix} = ab r(\cos^2\theta+\sin^2\theta) = ab r.

Thus $dA=ab r\,dr\,d\theta$. The factor $r$ accounts for the increasing circumference of circular slices as the radius grows, while $ab$ accounts for the stretch from a unit disk to the ellipse.

Elliptical-coordinate formula: For the upward orientation and a field satisfying the smoothness conditions, $$ \oint_{\partial S}\mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi}\int_0^1 \left[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right] (h+ar\cos\theta,\ k+br\sin\theta,\ z_0) \,ab r\,dr\,d\theta. $$ The derivatives are evaluated at the displayed point before integrating.

Why the Coordinate Conversion Works

Stokes’ Theorem equates the boundary circulation to a surface integral of the curl. On a horizontal patch with upward normal, the dot product with the normal selects the vertical curl component. The remaining step is an ordinary change of variables in a double integral.

In detail, the map $(r,\theta)\mapsto(x,y)$ covers the elliptical interior, with its center represented by $r=0$ and its boundary by $r=1$. The absolute Jacobian determinant gives the area multiplier $ab r$. Substituting the mapped coordinates into the vertical curl therefore gives exactly the displayed formula. No extra geometric factor is needed: the planar surface has $dS=dA$.

Worked Examples

Worked Example: Constant Curl on a Centered Ellipse

Let $\mathbf{F}=(-y,x,0)$, and let the boundary be the ellipse $x=2\cos\theta,\ y=\sin\theta,\ z=0$, traversed counterclockwise. Find its circulation using the elliptical interior, then check the result on the boundary.

Here $P=-y$ and $Q=x$, so

\frac{\partial Q}{\partial x}=1, \qquad \frac{\partial P}{\partial y}=-1, \qquad (\nabla\times\mathbf{F})_z=1-(-1)=2.

The ellipse has $a=2$, $b=1$, and center $(0,0)$. With $x=2r\cos\theta$, $y=r\sin\theta$, the curl-flux integral is

\begin{aligned} \oint_{\partial S}\mathbf{F}\cdot d\mathbf{r} &=\int_0^{2\pi}\int_0^1 2(2r)\,dr\,d\theta\\ &=\int_0^{2\pi}\left[2r^2\right]_0^1\,d\theta\\ &=\int_0^{2\pi}2\,d\theta\\ &=4\pi. \end{aligned}

To check, parameterize the boundary as $\mathbf{r}(\theta)=(2\cos\theta,\sin\theta,0)$. Then $dx=-2\sin\theta\,d\theta$ and $dy=\cos\theta\,d\theta$. Directly,

\begin{aligned} \oint_{\partial S}\mathbf{F}\cdot d\mathbf{r} &=\int_0^{2\pi}(-y\,dx+x\,dy)\\ &=\int_0^{2\pi} \left(2\sin^2\theta+2\cos^2\theta\right)\,d\theta\\ &=\int_0^{2\pi}2\,d\theta =4\pi. \end{aligned}

The direct boundary calculation agrees with the curl-flux calculation.

Worked Example: Variable Curl on a Translated Ellipse

Let $\mathbf{F}=(0,x^2,0)$, and consider the ellipse $x=1+2\cos\theta,\ y=-1+\sin\theta,\ z=0$, traversed counterclockwise. Its center is $(1,-1)$, and its semiaxis lengths are $a=2$ and $b=1$.

Since $P=0$ and $Q=x^2$, the vertical curl is

(\nabla\times\mathbf{F})_z = \frac{\partial(x^2)}{\partial x} - \frac{\partial 0}{\partial y} = 2x.

In scaled circular coordinates, $x=1+2r\cos\theta$, and the area factor is $2r$. Therefore,

\begin{aligned} \oint_{\partial S}\mathbf{F}\cdot d\mathbf{r} &=\int_0^{2\pi}\int_0^1 2(1+2r\cos\theta)(2r)\,dr\,d\theta\\ &=\int_0^{2\pi}\int_0^1 (4r+8r^2\cos\theta)\,dr\,d\theta\\ &=\int_0^{2\pi} \left(2+\frac{8}{3}\cos\theta\right)\,d\theta\\ &=4\pi+\frac{8}{3}\left[\sin\theta\right]_0^{2\pi}\\ &=4\pi. \end{aligned}

For a boundary check, $dy=\cos\theta\,d\theta$, so the line integral is

\begin{aligned} \oint_{\partial S}\mathbf{F}\cdot d\mathbf{r} &=\int_0^{2\pi}x^2\,dy\\ &=\int_0^{2\pi}(1+2\cos\theta)^2\cos\theta\,d\theta\\ &=\int_0^{2\pi} \left(\cos\theta+4\cos^2\theta+4\cos^3\theta\right)\,d\theta\\ &=0+4\pi+0\\ &=4\pi. \end{aligned}

Here the terms involving $\cos\theta$ and $\cos^3\theta$ integrate to zero over a full turn, while the $\cos^2\theta$ term contributes $\pi$.

Worked Example: A Quadratic Curl on an Ellipse

Let $\mathbf{F}=(0,x^3/3+xy^2,0)$, and use the counterclockwise ellipse $x=2\cos\theta,\ y=\sin\theta,\ z=0$. Evaluate the circulation with the interior and verify it along the boundary.

With $P=0$, the vertical curl is

(\nabla\times\mathbf{F})_z = \frac{\partial}{\partial x} \left(\frac{x^3}{3}+xy^2\right) = x^2+y^2.

Substitute $x=2r\cos\theta$, $y=r\sin\theta$, and $dA=2r\,dr\,d\theta$:

\begin{aligned} \oint_{\partial S}\mathbf{F}\cdot d\mathbf{r} &=\int_0^{2\pi}\int_0^1 \left(4r^2\cos^2\theta+r^2\sin^2\theta\right)2r\,dr\,d\theta\\ &=2\left(\int_0^1 r^3\,dr\right) \left(\int_0^{2\pi}(4\cos^2\theta+\sin^2\theta)\,d\theta\right)\\ &=2\left(\frac14\right)(4\pi+\pi)\\ &=\frac{5\pi}{2}. \end{aligned}

For the boundary check, $dy=\cos\theta\,d\theta$. Hence

\begin{aligned} \oint_{\partial S}\mathbf{F}\cdot d\mathbf{r} &=\int_0^{2\pi} \left(\frac{(2\cos\theta)^3}{3} +(2\cos\theta)\sin^2\theta\right)\cos\theta\,d\theta\\ &=\int_0^{2\pi} \left(\frac{8}{3}\cos^4\theta +2\cos^2\theta\sin^2\theta\right)\,d\theta\\ &=\frac{8}{3}\left(\frac{3\pi}{4}\right) +2\left(\frac{\pi}{4}\right)\\ &=2\pi+\frac{\pi}{2} =\frac{5\pi}{2}. \end{aligned}

The boundary integral and the interior integral give the same circulation.

A Common Pitfall: Omitting the Jacobian

A frequent error is to substitute the scaled coordinates into the curl but keep the area element $r\,dr\,d\theta$, as though the ellipse were a unit disk. The ellipse has been stretched by factors $a$ and $b$, so its area element must include $ab$.

For example, the first worked example has constant curl $2$ and semiaxes $a=2$, $b=1$. Omitting $ab$ would give the incorrect calculation

\int_0^{2\pi}\int_0^1 2r\,dr\,d\theta = 2\pi.

But the correct area factor is $ab r=2r$, so the correct integral is

\int_0^{2\pi}\int_0^1 2(2r)\,dr\,d\theta = 4\pi.

The missing factor would halve the answer. As a quick check for constant curl, the ellipse’s area is $\pi ab$, so a constant vertical curl $c$ must give circulation $c\pi ab$ for the upward orientation.

Orientation and a Reliable Setup

The coordinate formulas trace the boundary counterclockwise when $\theta$ increases from $0$ to $2\pi$. This matches the upward normal. If a problem specifies clockwise traversal, either reverse the sign of the counterclockwise result or use the downward normal when applying Stokes’ Theorem.

Identify the ellipse.
Read off its center, plane, and semiaxis lengths $a$ and $b$.

Match the orientation.
For counterclockwise traversal viewed from above, use the upward normal.

Compute the vertical curl.
For $\mathbf{F}=(P,Q,R)$, calculate $Q_x-P_y$.

Use the scaled coordinates.
Substitute $x=h+ar\cos\theta$, $y=k+br\sin\theta$, and multiply by $ab r$.

Check the sign and result.
Reverse the sign for the opposite orientation, and use a direct boundary integral when it is manageable.

This method is useful when the curl has a simpler expression than the field’s boundary integral, or when symmetry makes the interior integral easier to evaluate. It also makes the geometry explicit: the center determines where the curl is sampled, while the semiaxes determine the area scaling.

Check Your Understanding

Use the definitions and setup in this tutorial to answer the following questions.

For a counterclockwise elliptical boundary viewed from above, which unit normal is compatible with the orientation?
What are the bounds for $r$ and $\theta$ when covering the whole elliptical interior?
Why does the area element contain the factor $ab r$, rather than just $r$?
For $\mathbf{F}=(P,Q,R)$, which component of the curl contributes to flux through a horizontal patch with upward normal?
If a circulation calculation is performed for a clockwise traversal instead of a counterclockwise one, how does the answer change?