Let $S$ be a surface in $\mathbb{R^3} $ with the parametrization $g(s, t) = (t, t^2 , st)$ where $g : [0, 1] × [0, 10] → \mathbb{R^3} $ . Explain why the area of $S$ is equal to $\int_C xdσ$ , where $C$ is the curve in $\mathbb{R^2}$ parameterized by $h(t) = (t, t^2 )$, $h : [0, 10] → \mathbb{R^2}$ and Find the area of $S$.
i didn't understand the question at all how can i explain that . Area $S$ = $\int_C x\,d\sigma$ .
i know that the surface area is given by : $ \int f(x(t),y(t)\|r'(t)\| {\ dt}$
but the curve siting in the $[xy]$ plane is $h(t)$ so my guess (might be wrong) that $\|r'(t)\|$ = $\sqrt{4t^2 + 1}$ and $f(x(t),y(t)) = st = sx$ .
The question is only valid for this particular case, not in general.
Let me introduce a notation $\mathbf{g}(s,t)$ to denote the vector pointing from origin $(0,0,0)$ to the point $(t,t^2,st)$. Hence, $\mathbf{g}(s,t)=t\,\mathbf{i}+t^2\,\mathbf{j}+st\,\mathbf{k}$. The surface area is (refer any book on Advanced Engineering Mathematics) $$A=\int_0^{10}\int_0^1 \left\lVert \frac{\partial\mathbf{g}}{\partial s}\times\frac{\partial\mathbf{g}}{\partial t} \right\rVert\,\text{d}s\,\text{d}t = \int_0^{10}\int_0^1 \sqrt{4t^4+t^2}\,\text{d}s\,\text{d}t=\int_0^{10}t\sqrt{4t^2+1}\,\text{d}t.$$ Incidentally, in this case, $$\int_C x\,\text{d}\sigma=\int_0^{10}t\sqrt{4t^2+1}\,\text{d}t,$$ since $C$ is parametrised by $(t,t^2)$ and, as you already pointed out, the infinitesimal arc length is $\text{d}\sigma=\sqrt{4t^2+1}\,\text{d}t$.