Editing Torque (section)

==== Proof ====
The work done by a variable force acting over a finite linear displacement <math>s</math> is given by integrating the force with respect to an elemental linear displacement <math>\mathrm{d}\mathbf{s}</math>

<math display="block">W = \int_{s_1}^{s_2} \mathbf{F} \cdot \mathrm{d}\mathbf{s}</math>

However, the infinitesimal linear displacement <math>\mathrm{d}\mathbf{s}</math> is related to a corresponding angular displacement <math>\mathrm{d}\boldsymbol{\theta}</math> and the radius vector <math>\mathbf{r}</math> as 

<math display="block">\mathrm{d}\mathbf{s} = \mathrm{d}\boldsymbol{\theta}\times\mathbf{r}</math>

Substitution in the above expression for work, , gives
<math display="block">W = \int_{s_1}^{s_2} \mathbf{F} \cdot \mathrm{d}\boldsymbol{\theta} \times \mathbf{r}</math>

The expression inside the integral is a [[scalar triple product]] <math>\mathbf{F}\cdot\mathrm{d}\boldsymbol{\theta}\times\mathbf{r} = \mathbf{r} \times \mathbf{F} \cdot \mathrm{d}\boldsymbol{\theta}</math>, but as per the definition of torque, and since the parameter of integration has been changed from linear displacement to angular displacement, the equation becomes

<math display="block">W = \int_{\theta _1}^{\theta _2} \boldsymbol{\tau} \cdot \mathrm{d}\boldsymbol{\theta}</math>

If the torque and the angular displacement are in the same direction, then the scalar product reduces to a product of magnitudes; i.e., <math>\boldsymbol{\tau}\cdot \mathrm{d}\boldsymbol{\theta} = \left|\boldsymbol{\tau}\right| \left| \mathrm{d}\boldsymbol{\theta}\right|\cos 0 = \tau \, \mathrm{d}\theta</math> giving

<math display="block">W = \int_{\theta _1}^{\theta _2} \tau \, \mathrm{d}\theta</math>