# What are all the boundary curves for this combined cone and cylinder? [2013 10C]

Consider the bounded surface S that is the union of $x^2 + y^2 = 4$ for $−2 \le z \le 2$ and $(4 − z)^2 = x^2 + y^2$ for $2 \le z \le 4.$ Sketch the surface. Use suitable parametrisations for the two parts of S to verify Stokes’s Theorem for for $\mathbf{F} = (yz^2,0,0)$.

Herein, I ask only about computing line integrals instead of $\iint_S (\nabla × F )· d\mathbf{S}$, by virtue of Stokes's Theorem. I'm interested in an informal, intuitive answer. Denote the $2 \le z \le 4$ cone P, and the $-2 \le z \le 2$ cylinder C.

$1.$ What are all the boundary curves for this combined cone and cylinder? I think they are:
(i) $x^2 + y^2 \le 4$ on the $z = 2$ plane, while looking from above it?
(ii) $x^2 + y^2 \le 4$ on the $z = 2$ plane, while looking from below it?
(iii) I already consider here, though wrongly, (i) $x^2 + y^2 \le 4$ on the $z = -2$ plane, while looking from below it.

(iv) Any others that I've missed?

$2.$ Moreover, why does User Ellya's solution only consider (ii)?

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The first surface is a cylinder surface which is open at the top and bottom (the upper and lower disc of the cylinder is not part of the intended surface).

The second surface is a cone which is open at the botton (its bottom disc does not belong to the intended surface and its top point does not contribute, otherwise think of a tiny circle with radius $\epsilon$ at the top and then perform the limit $\epsilon \to 0$).

Their union is kind of a rocket / pointed stick, which is open at the bottom.

Therefore the only boundary of the rocket area is the curve

$$x^2 + y^2 = 4 \wedge z = -2$$

along the not included bottom disc.

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