$\Sigma M = 0$

$T = T_1 + T_2$ → Equation (1)

$\theta_1 = \theta_2$

$\left( \dfrac{TL}{JG} \right)_1 = \left( \dfrac{TL}{JG} \right)_2$

$\dfrac{T_1 a}{JG} = \dfrac{T_2 b}{JG}$

$T_1 = \dfrac{b}{a} T_2$ → Equation (2a)

$T_2 = \dfrac{a}{b} T_1$ → Equation (2b)

Equations (1) and (2b):

$T = T_1 + \dfrac{a}{b} T_1$

$T = \dfrac{T_1 b + T_1 a}{b}$

$T = \dfrac{(b + a) T_1}{b}$

$T = \dfrac{LT_1}{b}$

$T_1 = Tb / L$ (*okay!*)

Equations (1) and (2a):

$T = \dfrac{b}{a} T_2 + T_2$

$T = \dfrac{T_2 b + T_2 a}{a}$

$T = \dfrac{(b + a) T_2}{a}$

$T = \dfrac{L T_2}{a}$

$T_2 = Ta / L$ (*okay!*)

If the shaft were hollow, Equation (1) would be the same and the equality θ_{1} = θ_{2}, by direct investigation, would yield the same result in Equations (2a) and (2b). Therefore, the values of T_{1} and T_{2} are the same (*no change*) if the shaft were hollow.