The equal and opposite reaction [[Moment|torque]] can be interpreted as the sum of all internal [[Shear Stress|shearing stresses]]. $T=∫r\tau dA$ The following expression is obtained by integrating shear stress expressed with regards to shaft radius: $T=\tau_{max}\dfrac{\pi r^4}{2r_{max}}=\tau_{max}\dfrac{J}{r_{max}}$ Where $J$ is the polar moment of inertia, for circles: $J=\dfrac{\pi}{2}r^4$. >[!proof]- >$T=\int_A r\tau dA$ >$T=\int_0^r r\tau_{max}\dfrac{r}{r_{max}}2\pi rdr$ >$T=\tau_{max}\dfrac{2\pi r^4}{4r_{max}}$ >$T=\tau_{max}\dfrac{\pi r^4}{2r_{max}}$