Calculate $ \int \int_Sf(x,y,z)dS$
For part of the surface $x=z^3$, where $0\leq x,y \leq 14^{-\frac{3}{2}}$; $f(x,y,z) =x$
The parametrization of the surface seems simple enough:
$s(u,v) = \langle u^3,v,u \rangle$
Thus we compute the normal to the surface:
$s_u = \langle 3u^2,0,1\rangle$; $s_v = \langle 0,1,0 \rangle$
The jacobian yields:
$s_u \times s_v = \langle 1,0,3u^2\rangle$
$|s_u \times s_v| = \sqrt{9u^4+1}$
Ans thus:
$\int \int_Sf(x,y,z)dS = \int \int_Df(s(u,v))\|\vec n\|dA = \int^{14^{-3/2}}_0 \int^{14^{-3/2}}_0 u^3\sqrt{9u^4+1}$ $dudv$
Which according to Wolfram is $\approx 6.338403326748645^{-10}$
This is wrong, and I am not sure why it is wrong.
$u$ represents $z$ and $x=z^3=u^3$,
If $0 \leq x \leq 14^{-\frac32}$, we should have
$$0 \leq u \leq 14^{-\frac12}.$$