For $\beta=1/2$, the exact profiles are
$$
T=T_0(1-x^2)^{1/2},
\qquad
\rho=\rho_0(1-x^2),
\qquad
P=P_0(1-x^2)^2,
\qquad x=\frac zH.
$$
Integration through both disk faces gives
$$
\boxed{\Sigma=\rho_0H\int_{-1}^1(1-x^2)\,dx
=\frac43\rho_0H}.
$$
Since $\mu=\alpha P/\Omega$ and $\bar\nu\Sigma=\int\mu\,dz$,
$$
\boxed{\bar\nu\Sigma
=\frac{\alpha P_0H}{\Omega}
\int_{-1}^1(1-x^2)^2\,dx
=\frac{16}{15}\frac{\alpha P_0H}{\Omega}}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 321 / 1 / c / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-321/1/c/i/solution