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For a continuously differentiable vector field on an oriented surface with compatibly oriented boundary, this vector identity follows by dotting both sides with an arbitrary constant vector $a$ and applying Stokes theorem to $\mathbf{F}\times a$. The curl identity $\nabla \times (a\times \mathbf{F})=a\nabla \cdot \mathbf{F}-(a\cdot \nabla )\mathbf{F}$ gives the surface integrand. In components, $[(d\mathbf{S}\times \nabla )\times \mathbf{F}{]}_i=d{S}_j{\partial }_i{F}_j-d{S}_i{\partial }_j{F}_j$, with the derivatives acting on $\mathbf{F}$. For the position vector field $\mathbf{F}=\mathbf{r}$, it equals $-2d\mathbf{S}$; hence the line integral is minus twice the oriented vector area.

A double pendulum consists of two pendula coupled because the second pivot moves with the first mass. It has two degrees of freedom: even when the upper pivot is fixed, moving the upper arm carries the lower pivot along. This makes the double pendulum a compact example of how mechanical constraints couple motion, and how Lagrangian mechanics converts geometry into dynamics.

Use two equal point masses $m>0$ and two massless rigid rods of equal length $\ell >0$, with joints that allow free rotation in one vertical plane. The upper pivot is fixed; gravity is uniform, with gravitational acceleration $g>0$. Neglect joint friction and drag force. Both angles ${\theta }_1,{\theta }_2$ are measured from the downward vertical, positive toward the right, in radians. In particular, ${\theta }_2$ is an absolute angle, not the bend between the rods: that bend is ${\theta }_2-{\theta }_1$.

The double pendulum Lagrangian below gives the exact dynamics of this ideal model. Near the downward stable equilibrium, normal modes of a double pendulum replace the coupled motion with two independent harmonic oscillators. Their fixed amplitude ratios explain which releases make the rods swing together and which make them swing in opposite directions.

Choose Cartesian coordinates with origin at the upper pivot, $x$ to the right and $y$ upward. The two position vectors areThese expressions satisfy the rod holonomic constraints $|{\mathbf{r}}_1|=\ell$ and $|{\mathbf{r}}_2-{\mathbf{r}}_1|=\ell$ automatically. This is the practical advantage of using the two angles as coordinates: the rod tensions need not be unknowns in the motion calculation.^[ref]

Differentiate these position vectors to obtain the velocities. The dot product between their two rod contributions uses $\cos ({\theta }_1-{\theta }_2)=\cos {\theta }_1\cos {\theta }_2+\sin {\theta }_1\sin {\theta }_2$, givingAdding the two kinetic energies and measuring gravitational potential energy from the downward equilibrium givesThe coefficient $2$ has a geometric meaning: the upper rod raises both masses, and its motion contributes to both kinetic energies. The coupling in $T$ is present because the lower mass has a moving pivot. The zero chosen for $V$ changes $L$ by a constant and leaves the Euler-Lagrange equations unchanged.^[ref]

The principle of stationary action applies to $S=\int Ldt$. Its fixed-endpoint first variation, after integration by parts, isIndependent allowed variations therefore give one Euler-Lagrange equation for each angle. Stationary action alone does not assert a minimum.^[ref]

Let a solid cone of volume $V$ have its planar base in a plane through the origin, and orient its curved surface $C$ outwards. The vector field $\mathbf{F}(\mathbf{x})=\mathbf{x}$ has divergence three. Its flux through the base vanishes because the position vector lies in that plane and is orthogonal to its normal vector. The divergence theorem therefore gives the displayed curved-surface flux. For a circular base of radius $a$ and perpendicular height $b$, this is $\pi a^{2}b$.

The position vectors end at the midpoints of $BC$, $CA$ and $AB$, respectively. The drawing places $E$, $F$ and $G$ on those edges and shows the arrows from the origin. The volume of their parallelepiped is the absolute scalar triple product:Equivalently, this is the absolute determinant of the matrix whose columns are the three vectors. Since it is nonzero, the vectors are linearly independent. Three linearly independent vectors in the three-dimensional vector space ${\mathbb{R}}^3$ form a basis.

Let $\mathbf{d}=\mathbf{f}-\mathbf{e}$. A point on the line has position vector $\mathbf{r}=\mathbf{e}+t\mathbf{d}$. By bilinearity and antisymmetry of the cross product,For a point with position vector $\mathbf{c}$, the area of the parallelogram spanned by $\mathbf{c}-\mathbf{e}$ and $\mathbf{d}$ equals base length times perpendicular Euclidean distance. Hence the shortest Euclidean distance to the line isFor $\mathbf{c}=(0,0,1)$, these quantities areThus $\boxed{d=\sqrt{6}/4}$. The perpendicular meets $EF$ at $(1/4,1/4,1/2)$, whose displacement from $C$ has precisely this length.

For a twice continuously differentiable vector field in Cartesian coordinates, the curl and divergence arePut $\mathbf{\omega }=\nabla \times \mathbf{v}$. Using the Levi-Civita symbol, the Kronecker delta contraction and the product rule,The last term vanishes because the divergence of a curl is zero: the symmetric mixed derivatives contract against an antisymmetric symbol. Thus the curl of a cross product identity becomes

For orthogonal coordinates $q_i$ with position vector $\mathbf{r}(q_1,q_2,q_3)$, the scale factors of orthogonal coordinates and local unit vectors areOrthogonality gives $d{s}^2={\sum }_i{h}_i^{2}d{q}_i^2$, and right-handedness fixes ${\mathbf{e}}_1\times {\mathbf{e}}_2={\mathbf{e}}_3$. In cylindrical coordinates, $\mathbf{r}=(\rho \cos \varphi ,\rho \sin \varphi ,z)$, soThe radial vector points away from the axis; the azimuthal vector is tangent in the direction of increasing $\varphi$; ${\mathbf{e}}_z$ is the Cartesian positive $z$ direction. The azimuthal basis is undefined on the coordinate axis itself.

For the axial velocity profile, the cylindrical coordinates formula givesThe associated cross product isThis field has only a radial component depending on $\rho$, so its curl is zero. Also $\nabla \cdot \mathbf{u}={\partial }_z{u}_z=0$. The identity above therefore yieldsIn fact both directional derivatives vanish: the first differentiates $u_z{\mathbf{e}}_z$ only in the azimuthal direction, and the second differentiates $\mathbf{\omega }$ only in the $z$ direction. All physical vector fields extend smoothly through $\rho =0$, despite the coordinate-basis singularity.

Using Cartesian coordinates, the Levi-Civita symbol and the Einstein summation convention, the constant-vector assumption givesRecognizing the divergence and directional derivative provesHere the contractions use the Kronecker delta.

For an oriented piecewise smooth surface $S$ with boundary $C$, Stokes theorem stateswhere the orientations obey the right-hand rule and $\mathbf{Q}$ is continuously differentiable near $S$. Apply it to $\mathbf{Q}=\mathbf{F}\times \mathbf{a}$. The scalar triple product givesThe last integrand is $\mathbf{a}\cdot [(d\mathbf{S}\times \nabla )\times \mathbf{F}]$, as the component formulashows. Since $\mathbf{a}$ is arbitrary, the cross-product form of Stokes theorem follows:The derivatives in the surface expression act on $\mathbf{F}$, not on the surface element.

For the specified square, the path is clockwise when viewed from positive $z$, so the compatible unit normal is $-{\mathbf{e}}_z$. On the first and fourth edges, $d\mathbf{l}$ and $\mathbf{r}$ are parallel, giving zero cross product. On the second edge, $y=1$ and $x$ runs from $0$ to $1$, so $d\mathbf{l}\times \mathbf{r}={\mathbf{e}}_{z}dx$ contributes ${\mathbf{e}}_z$. On the third edge, $x=1$ and $y$ runs from $1$ to $0$, so the contribution is $-{\mathbf{e}}_z{\int }_1^{0}dy={\mathbf{e}}_z$. Thus the line integral is $2{\mathbf{e}}_z$.

For an independent surface integral, ${\partial }_i{F}_j={\delta }_{ij}$ and $\nabla \cdot \mathbf{F}=3$ for the position vector field. Its integrand is therefore $d\mathbf{S}-3d\mathbf{S}=-2d\mathbf{S}$. The square has unit area and downward orientation, giving

In the usual physics convention for spherical coordinates, the polar angle is the angle between the positive $z$ axis and the position vector. For $r=\sqrt{x^2+y^2+z^2}>0$, it satisfies $\cos \theta =z/r$ with $0\le \theta \le \pi$. The north and south axes have $\theta =0$ and $\theta =\pi$, respectively; the plane $z=0$ has $\theta =\pi /2$. The angle is undefined at the origin. In plane polar coordinates, the same phrase usually denotes the oriented in-plane angle instead, so the reference axis and convention should always be stated. The azimuth supplies the other angular coordinate in three dimensions.

A position vector is the vector from a chosen origin to a point. Its time derivative is the velocity; differences of position vectors give displacement.

Let $D=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})>0$. The reciprocal basis gives the coordinates of any position vector asThe point lies in the closed tetrahedron with vertices $0,\mathbf{a},\mathbf{b},\mathbf{c}$ exactly when $\alpha ,\beta ,\gamma \ge 0$ and $\alpha +\beta +\gamma \le 1$. Indeed, these conditions express it as a convex combination of the four vertices, with fourth weight $1-\alpha -\beta -\gamma$. They also describe the intersection of the four inward half-spaces bounded by its faces. Strict inequalities characterize its interior.