I just want to confirm that my calculation is right with the following problem.
I want to find the rate of heat flow where $u(x,y,z) = 2y^2+2z^2$ , $K=6.5$ through surface $S$... $y^2+z^2=6, \space x\in[0,4]$
I know that
$$\int \int_S -K\nabla \vec u\cdot d\vec S$$
is the usual heatflow equation, but the problem asks us to find the heatflow inwards the surface so the answer will simply be the opposite of this expression.
Without nice shortcuts or anything, I parametarized the surface using $y = \sqrt{6}\cos{\theta}$, $z = \sqrt{6}\sin{\theta}$ and $x=x$ so
$$\nabla \vec u = <0,4\sqrt{6}\sin{\theta},4\sqrt6\cos{\theta}>$$
and
$$\vec n = <0,\sqrt6\cos{{\theta},} \sqrt{6} \sin{\theta}>$$
where the dot product is simply $24$.
So, the inflow simply becomes
$$K(24)A$$
where $A$ is the lateral area of the cylinder, which is $8\sqrt{6}\pi$ .
I get the solution $1248\sqrt{6}\pi$, but can anyone else confirm that I got the right number?
$\nabla \vec u = <0,4\sqrt{6}\cos{\theta},4\sqrt6\sin{\theta}>$. You just have a typo as your reversed the order of $y$ and $z$ component when you wrote but your dot product is correct.
$\vec n = \ <0, \sqrt6\cos{{\theta},} \sqrt{6} \sin{\theta}>$ is radially outward vector but given the question we will choose $\vec n = <0, - \sqrt6\cos{{\theta},} - \sqrt{6} \sin{\theta}>$.
If you do your calculation using outward pointing vector, you will have to just reverse the sign in the end as you rightly said.
Now this is where you have a mistake. Your normal vector is not normalized so multiplying again by surface area will give a wrong result.
Since you have parametrized the surface as $r(x,\theta) = <x, \sqrt6 \cos\theta, \sqrt6 \sin\theta>$
$dS = \vec{n}\ dx \ d\theta $ ($\vec{n} = r'_{x} \times r'_{\theta}$). So your integral is simply,
$\displaystyle 24K \int_0^{2\pi} \int_0^{4} dx \ d\theta = 1248 \pi$