Let $E=\{(x,y):x^2+y^4\leq 16\}$ and $S$ the surface in $\mathbb R^3$ which is the graph of the function $f(x,y)=16-x^2-y^4$ and let $\Omega$ be the region enclosed by $S$ and $\hat{E}=\{(x,y,0):(x,y)\in E\}$. Let $v$ be a vector field define by $v(x,y,z)=(2x,y,4z)$
I'm asked to use the divergence theorem to determine a relationship between the volume of $\Omega$ and the area of $E$. So by the divergence theorem we know that, since $\nabla\cdot v=7$ $$\iiint_{\Omega}7\,dV=\iint_Sv\cdot n\,ds+\iint_{\hat{E}}v\cdot n\,ds$$ For the integral on the right we can parameterise $\hat{E}$ by $r(t)=(4\cos t,4\sin t,0)$. Clearly the normal pointing out is given by $n=(0,0,-1)$ and so the integral on the right is $0$.
For the left integral, parameterising $S$ by $r(x,y)=(x,y,16-x^2-y^4)$ and $n$ is given by $n=(2x,4y^3,1)$. Then $$\iint_Sv\cdot n\,ds=\iint_S(2x,y,64-4x^2-4y^4)\cdot(2x,4y^3,1)\,dA=\iint_S64\,dA=64Area(S)$$
So $Vol(\Omega)=\frac{64}{7}Area(S)$, but the question asks for the volume in terms of the area of $E$. Have I gone wrong or is it likely that it meant $S$?
In the solution they say $$\iint_Sv\cdot n\,ds=\iint_E(2x,y,64-4x^2-4y^4)\cdot(2x,4y^3,1)\,dA$$ How?
This is how. Note that the integration at the end is over $T$ (corresponding to your $E$) with respect to $ds\,dt$ which is an area differential in $T$.
The $n$ you produced is not the unit normal to $S$. Note that for most $x,y$, it will not be a unit vector. What you have is the normal vector with the Jacobian for area conversion from $E$ to $S$ built-in.