= Solution
Take the <normal vector> outward from the cylinder-and-cone solid. The surface consists of the cylindrical side and conical roof, with no bottom disk; this convention fixes the otherwise unspecified <orientation of a surface>. The boundary is the radius-two circle at $z=-2$. Reversing every <normal vector> reverses the final sign.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/ia/paper-3-cylinder-cone.png]
{title=Outward-oriented cylindrical side and conical roof, with the open bottom circle as the sole boundary}
For $\mathbf F=(yz^2,0,0)$, its <curl> is $(0,2yz,-z^2)$. For the cylindrical <parametrized surface>, use $\mathbf r_c=(2\cos\theta,2\sin\theta,z)$, with $-2\le z\le2$, and use $\mathbf r_{c,\theta}\times\mathbf r_{c,z}=(2\cos\theta,2\sin\theta,0)$ as the outward area vector. Its flux is
$$
I_c=\int_{-2}^{2}\int_0^{2\pi}8z\sin^2\theta\,d\theta\,dz=0.
$$
For the cone, let $q=4-z$, $2\le z\le4$ and $\mathbf r_k=(q\cos\theta,q\sin\theta,z)$. Its outward area vector is $\mathbf r_{k,\theta}\times\mathbf r_{k,z}=(q\cos\theta,q\sin\theta,q)$. Thus
$$
\begin{aligned}
I_k&=\int_2^4\int_0^{2\pi}(2q^2z\sin^2\theta-qz^2)\,d\theta\,dz\\
&=2\pi\int_2^4(16z-12z^2+2z^3)\,dz=-16\pi.
\end{aligned}
$$
The total <surface integral> is therefore \b[$-16\pi$ with the outward convention], or $+16\pi$ for the opposite convention.
For <Stokes theorem>, the induced bottom-circle orientation is counterclockwise as viewed from above, opposite to the orientation of an outward bottom cap. Use $\mathbf r=(2\cos\theta,2\sin\theta,-2)$ with increasing $\theta$. Then
$$
\boxed{\oint_{\partial S}\mathbf F\cdot d\mathbf r
=\int_0^{2\pi}(8\sin\theta)(-2\sin\theta)\,d\theta=-16\pi,}
$$
agreeing with the two parametrized fluxes. The seam at $z=2$ is internal and cancels between the two pieces; the cone tip does not supply another boundary curve.
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