The surface $S$ is defined as the set of all points in $\mathbb{R}^3$ whose coordinates satisfy $x^2 + y^2 = z^2$.
What is the image of your curve? Well, the points of the image of your curve are precisely points of the form $r(t) = (\sin t\cos t, \cos^2 t, \cos t)$. As $t$ ranges over all of $\mathbb{R}$, $r(t)$ ranges over all points in the image of your curve.
Thus, you want to ask: Do all points in the image of your curve have coordinates satisfying $x^2 + y^2 = z^2$? The important thing to note is that here $x,y,z$ just refer to the first, second, and third coordinates of your points. Well, the points of the image of your curve are of the points with coordinates $(\sin t\cos t,\cos^2 t,\cos t)$, and thus you want to check if the coordinates satisfy:
(first coordinate squared) + (second coordinate squared) = (third coordinate squared)
How do you check that? Well you simply verify that
$$(\sin t\cos t)^2 + (\cos^2 t)^2 = (\cos t)^2$$
for all $t\in\mathbb{R}$.
If this is true, then that's to say that for any $t\in\mathbb{R}$, the point $(\sin t\cos t,\cos^2 t,\cos t)$ satisfies (first coordinate squared) + (second coordinate squared) = (third coordinate squared), and thus lies on your surface $S$.
Since every point on your curve can be described in this way, you're done.
The parametrization does not correspond to the cap. Try this parametrization.
$$r(u, v) = (u\cos v, u\sin v, \sqrt{2-u^2})$$
Then
$$r_u = \left(\cos v, \sin v, \frac{-u}{\sqrt{2-u^2}}\right)$$
and
$$r_v = (-u\sin v, u\cos v, 0)$$
One can see geometrically and algebraically that $r_u$ is perpendicular to $r_v$. Thus, the surface element is
$$\left\lVert \frac{dr}{du} \times \frac{dr}{dv}\right\rVert = \left\lVert \frac{dr}{du} \right\rVert \left\lVert \frac{dr}{dv}\right\rVert = u\sqrt{1 + \frac{u^2}{2-u^2}} = \frac{\sqrt{2}u}{\sqrt{2-u^2}}$$
Thus, the surface area is
$$\iint_{[0,1] \times [0, 2\pi]} \frac{\sqrt{2} u}{\sqrt{2-u^2}} \: dudv = \sqrt{2} \int_{0}^{2\pi} \Big[\sqrt{2-u^2}\Big]_1^0 \: dv = \sqrt{2}(\sqrt{2}-1)2\pi = \boxed{(4 - 2\sqrt{2})\pi}$$
Best Answer
Yes, but it is not the (four-dimensional) graph of the function $f:\mathbb{R}^3\to\mathbb{R}$ defined by $$f(x,y,z)=\frac{\sin(\sqrt{x^2+y^2+z^2})}{\sqrt{x^2+y^2+z^2}},$$ but rather a slice of it; that is, choose some value $z=c$, and this graph is the graph of the function $g:\mathbb{R}^2\to\mathbb{R}$ defined by $$g(x,y)=\frac{\sin(\sqrt{x^2+y^2+c^2})}{\sqrt{x^2+y^2+c^2}}.$$ Choosing different values of $c$ varies the graph; here is an animation of the graph of $g$ as $c$ goes from $0$ to $8$:
listofplots = {}; For[c = 0, c <= 40, c++, AppendTo[listofplots, Plot3D[Sin[Sqrt[x^2 + y^2 + (c/5)^2]]/ Sqrt[x^2 + y^2 + (c/5)^2], {x, -15, 15}, {y, -15, 15}, PlotPoints -> 40, PlotRange -> {-0.2, 0.2}]]] Export["animation.gif", listofplots, "DisplayDurations" -> {0.25}]