Solid-Body Rotation and Stokes’ Theorem

Rigid Rotation as a Velocity Field

A rigid body can translate, rotate, or do both at once. In a rigid motion, every point has the same translational velocity, while the rotational part depends on the point’s position relative to a reference point. The resulting field gives a concrete setting in which to compare local rotation, measured by curl, with total circulation along a boundary.

Let $\mathbf r_0$ be a fixed reference position, let $\mathbf U$ be its translational velocity, and let $\boldsymbol\Omega$ be a constant angular-velocity vector. The velocity at position $\mathbf r$ is

\mathbf v(\mathbf r) = \mathbf U+\boldsymbol\Omega\times(\mathbf r-\mathbf r_0).

Definition: A rigid-body velocity field has the form $\mathbf v(\mathbf r)=\mathbf U+\boldsymbol\Omega\times(\mathbf r-\mathbf r_0)$, where $\mathbf U$, $\boldsymbol\Omega$, and $\mathbf r_0$ are constant. The term $\mathbf U$ describes translation, and the cross-product term describes rotation.

This field is affine in position: it is a linear function of $\mathbf r-\mathbf r_0$, followed by addition of the constant vector $\mathbf U$. It is not generally a linear map of $\mathbf r$, since a linear map must send the zero vector to zero. In particular, $\mathbf v(\mathbf r_0)=\mathbf U$, which need not vanish. When $\mathbf U=\mathbf 0$, the field is linear in $\mathbf r-\mathbf r_0$.

The Curl of Rigid Rotation

The curl measures local rotational tendency in a vector field. For a constant angular velocity, the curl of the rotational part is itself constant. To see this directly, write $\boldsymbol\Omega=(a,b,c)$ and $\mathbf r=(x,y,z)$. The constant reference point only contributes constants to the components of the field, and constants have zero partial derivatives. The variable part of the cross product is

\boldsymbol\Omega\times\mathbf r = (bz-cy,\;cx-az,\;ay-bx).

Using the component definition of curl, each component can be computed explicitly:

\begin{aligned} \bigl(\nabla\times\mathbf v\bigr)_x &= \frac{\partial v_z}{\partial y} - \frac{\partial v_y}{\partial z} = a-(-a)=2a,\\ \bigl(\nabla\times\mathbf v\bigr)_y &= \frac{\partial v_x}{\partial z} - \frac{\partial v_z}{\partial x} = b-(-b)=2b,\\ \bigl(\nabla\times\mathbf v\bigr)_z &= \frac{\partial v_y}{\partial x} - \frac{\partial v_x}{\partial y} = c-(-c)=2c. \end{aligned}

Thus the translation contributes nothing to the curl, and the rotational part gives

\nabla\times\mathbf v=2\boldsymbol\Omega.

The factor of two is important: for rotation about the positive $z$-axis with angular speed $\omega$, the planar velocity field is $(-\omega y,\omega x,0)$, whose curl points in the positive $z$-direction and has magnitude $2\omega$.

Stokes’ Theorem and Orientation

For a smooth vector field $\mathbf v$ defined on an open neighborhood of an oriented, piecewise-smooth surface $S$, Stokes’ theorem relates the circulation along its boundary to the flux of curl through the surface. The direction around the boundary must agree with the chosen surface normal by the right-hand rule.

Theorem: If $S$ is an oriented surface with positively oriented boundary curve $\partial S$, and $\mathbf v$ has continuous first partial derivatives on a neighborhood of $S$, then $$ \oint_{\partial S}\mathbf v\cdot d\mathbf r = \iint_S(\nabla\times\mathbf v)\cdot\mathbf n\,dS. $$ Here $\mathbf n$ is the unit normal consistent with the orientation of the boundary.

For a planar region of area $A$ with constant unit normal $\mathbf n$, rigid-body rotation gives an especially direct result. Since the curl is the constant vector $2\boldsymbol\Omega$, its component normal to the plane is constant across the region. Therefore Stokes’ theorem gives

\oint_{\partial S}\mathbf v\cdot d\mathbf r = \iint_S2\boldsymbol\Omega\cdot\mathbf n\,dS = 2(\boldsymbol\Omega\cdot\mathbf n)A.

The formula includes translation: its circulation around any closed curve is zero, because the integral of the constant vector $\mathbf U$ around a closed path is $\mathbf U\cdot\oint_{\partial S}d\mathbf r=\mathbf 0$. The rotational contribution depends on the component of angular velocity perpendicular to the plane and on the enclosed area.

Worked Example: Circular Boundary

Worked Example: Circulation Around a Rotating Disk

Consider $\mathbf v(x,y,z)=(-3y,3x,0)$ and the circle $x^2+y^2=4$ in the $xy$-plane, oriented counterclockwise as viewed from the positive $z$-axis. Compute its circulation directly, then verify it with Stokes’ theorem.

Here $\boldsymbol\Omega=(0,0,3)$, and a positive parametrization of the circle is $\mathbf r(t)=(2\cos t,2\sin t,0)$, for $0\le t\le2\pi$. Its derivative and the velocity along the path are

\mathbf r'(t)=(-2\sin t,2\cos t,0), \qquad \mathbf v(\mathbf r(t))=(-6\sin t,6\cos t,0).

The dot product is $\mathbf v(\mathbf r(t))\cdot\mathbf r'(t)=12\sin^2t+12\cos^2t=12$. Thus the direct circulation is

\oint_{\partial S}\mathbf v\cdot d\mathbf r = \int_0^{2\pi}12\,dt = 24\pi.

For the Stokes calculation, $\nabla\times\mathbf v=(0,0,6)$, the upward unit normal is $\mathbf n=(0,0,1)$, and the disk area is $\pi(2)^2=4\pi$. Hence

\iint_S(\nabla\times\mathbf v)\cdot\mathbf n\,dS = \int_0^{4\pi}6\,dS = 24\pi.

The two calculations agree. The direct integrand is constant, and the curl flux is the curl magnitude times the disk area.

Worked Example: Elliptical Boundary

Worked Example: Circulation Around an Ellipse

Let $\mathbf v(x,y,z)=(-2y,2x,0)$. Find the circulation around the ellipse $x^2/9+y^2/4=1$ in the $xy$-plane, oriented counterclockwise. Verify the answer using Stokes’ theorem.

Parametrize the ellipse by $\mathbf r(t)=(3\cos t,2\sin t,0)$, $0\le t\le2\pi$. Then

\mathbf r'(t)=(-3\sin t,2\cos t,0), \qquad \mathbf v(\mathbf r(t))=(-4\sin t,6\cos t,0).

The dot product is

\mathbf v(\mathbf r(t))\cdot\mathbf r'(t) = 12\sin^2t+12\cos^2t = 12.

Consequently, the direct line integral is $\int_0^{2\pi}12\,dt=24\pi$. For the surface calculation, the curl is $(0,0,4)$. The ellipse has semiaxes $3$ and $2$, so its area is $3\cdot2\cdot\pi=6\pi$. With upward normal, Stokes’ theorem gives

\iint_S(\nabla\times\mathbf v)\cdot\mathbf n\,dS = 4(6\pi)=24\pi.

The direct result and the curl flux agree. The boundary is not circular, but its circulation still follows the same area rule for this constant-curl field.

Worked Example: Rectangular Boundary with Translation

Worked Example: Translation Does Not Change Closed-Path Circulation

Consider $\mathbf v(x,y,z)=(2-y,x-1,0)$ and the rectangle with vertices $(0,0)$, $(4,0)$, $(4,3)$, and $(0,3)$, traversed in that order. Compute the circulation along all four sides and verify it by Stokes’ theorem.

The field consists of a translation $(2,-1,0)$ and rotation with $\boldsymbol\Omega=(0,0,1)$. Traverse the four sides counterclockwise. On the bottom edge, take $x=t,y=0$, for $0\le t\le4$. Then $d\mathbf r=(1,0,0)\,dt$, so $\mathbf v\cdot d\mathbf r=2\,dt$, and the contribution is $\int_0^4 2\,dt=8$.

On the right edge, take $x=4,y=t$, for $0\le t\le3$. Here $d\mathbf r=(0,1,0)\,dt$, and the second component of the velocity is $x-1=3$. The contribution is $\int_0^3 3\,dt=9$.

On the top edge, take $x=t,y=3$, with $t$ decreasing from $4$ to $0$. The first velocity component is $2-3=-1$, so $\mathbf v\cdot d\mathbf r=-dt$. This edge contributes $\int_4^0(-1)\,dt=4$.

On the left edge, take $x=0,y=t$, with $t$ decreasing from $3$ to $0$. The second velocity component is $x-1=-1$, so this edge contributes $\int_3^0(-1)\,dt=3$. Adding the four contributions gives $8+9+4+3=24$.

The curl is $(0,0,2)$, the upward normal is $(0,0,1)$, and the rectangle’s area is $4\cdot3=12$. Stokes’ theorem gives $2\cdot12=24$, confirming the direct calculation. The translation has nonzero contributions on individual sides, but its total contribution around the closed rectangle is zero.

A Derivation of the Area Rule

The circulation-area formula can be understood by combining the constant-curl calculation with Stokes’ theorem. Suppose $S$ is a planar region of area $A$, with unit normal $\mathbf n$, and its boundary orientation is positive. Rigid-body velocity is smooth everywhere, so it meets the differentiability requirement of Stokes’ theorem. At every point of $S$, the curl is $2\boldsymbol\Omega$. Therefore the integrand is constant:

(\nabla\times\mathbf v)\cdot\mathbf n = (2\boldsymbol\Omega)\cdot\mathbf n = 2(\boldsymbol\Omega\cdot\mathbf n).

Integrating a constant over a region multiplies that constant by the region’s area. Applying Stokes’ theorem now gives

\oint_{\partial S}\mathbf v\cdot d\mathbf r = \iint_S2(\boldsymbol\Omega\cdot\mathbf n)\,dS = 2(\boldsymbol\Omega\cdot\mathbf n)\iint_SdS = 2(\boldsymbol\Omega\cdot\mathbf n)A.

This derivation explains why the shape of a planar boundary does not appear in the final formula: for this field, curl is constant, so only the area and the orientation of the region matter. A tilted region uses only the component of $\boldsymbol\Omega$ along its normal.

Common Pitfalls: Linearity and Orientation

A tempting but incorrect description is to call $\mathbf v(\mathbf r)=\mathbf U+\boldsymbol\Omega\times(\mathbf r-\mathbf r_0)$ a linear function of position without qualification. For example, if $\mathbf U\ne\mathbf0$, then $\mathbf v(\mathbf r_0)=\mathbf U\ne\mathbf0$, contradicting the defining property of a linear map that the zero input maps to zero. The correction is to call the field affine in $\mathbf r$; its rotational part is linear in $\mathbf r-\mathbf r_0$.

Orientation is another source of sign errors. In the circular example, the counterclockwise path viewed from above agrees with the upward normal and gives circulation $24\pi$. Reversing the path changes the line integral to $-24\pi$. If the upward normal is kept fixed while applying Stokes’ theorem, its flux remains $24\pi$, so the orientations no longer match. To use Stokes’ theorem correctly, reverse the normal as well, or restore the original boundary direction.

Key takeaway: For a rigid-body velocity field, $\nabla\times\mathbf v=2\boldsymbol\Omega$. Around a positively oriented planar boundary of area $A$, the circulation is $2(\boldsymbol\Omega\cdot\mathbf n)A$. Translation has zero net circulation around a closed curve.

Check Your Understanding

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

Why is a rigid-body velocity field with nonzero translation generally affine rather than linear in position?
What is the curl of $\mathbf v(\mathbf r)=\mathbf U+\boldsymbol\Omega\times(\mathbf r-\mathbf r_0)$?
For a planar region with unit normal $\mathbf n$ and area $A$, what determines the sign of its circulation?
Why does a constant translational velocity contribute zero circulation around a closed boundary?
If a positively oriented boundary is traversed in reverse, what must change in the Stokes calculation?