Answer
$1248 \pi$
Work Step by Step
Since, $\iint_S F \cdot dS=\iint_S F \cdot n dS$
where, $n$ denotes the unit vector.
and $dS=n dA=\pm \dfrac{(r_u \times r_v)}{|r_u \times r_v|}|r_u \times r_v| dA=\pm (r_u \times r_v) dA$
The flux through a surface can be defined only when the surface is orientable.
$r_s \times r_v=\sqrt 6 (\cos u j+\sin u k)$
and $dS=\sqrt 6 ( \cos u j+\sin u k) dA$
Here, $F(r(u,v))=26 \sqrt 6 (\cos u j+\sin u k)$
$\iint_S F \cdot dS= \int_{0}^4 \int_{0}^{2 \pi} 26 \sqrt 6 (\cos u j+\sin u k) [\sqrt 6 ( \cos u j+\sin u k) dA] $
$=\int_{0}^4 \int_{0}^{2 \pi} 26(6) (\cos^2 u+\sin^2 u) dA$
$=(26) \cdot (6) \cdot (4) \cdot (2 \pi)$
Hence, the rate of heat inflow is $1248 \pi$
