Given two parameterized functions of variable $t\in \mathbb{R}$: $$ \begin{cases} \vec{f}: \mathbb{R}\longrightarrow \mathbb{R}^3 \\ \vec{f}(t)=(x(t),y(t),z(t)) \end{cases} \quad \quad \begin{cases} \vec{g}: \mathbb{R}\longrightarrow \mathbb{R}^2 \\ \vec{g}(t) = (x'(t),y'(t)) \end{cases} $$ of which both are continuous on their domain. Also with the projection $$proj(\vec{f})_{xy}=C(x(t),y(t),0)$$ on the $xy$-plane, the following limit is true: \begin{align} \lim_{ t' \to +\infty } \int_{C[x(t'),y(t'),0]} \hat{i} x(t)+\hat{j} y(t) +\hat{k} z(t) \, ds \neq \pm\infty \\ \lim_{ t' \to -\infty } \int_{C[x(t'),y('t'),0]} \hat{i} x(t)+\hat{j} y(t) +\hat{k} z(t) \, ds \neq \pm\infty \end{align} Additionally, $C\neq g$, that is two functions do not intersect at any point.
- General solution to find the area of the surface $S$ that intersects both $f$ and $g$?
- General solution to calculatethe volume under $S$?
What is the solution for this question? I cannot solve this, but I am trying to look for the geometric perspective. The wording is also quite hard to understand, and maybe redundant, because the limits given is essentially saying that the line integral is finite. Regardless, I did tried with $$\vec{f}(t)=\left(2t^2, \frac{4}{3}t + \sqrt{t}, 5t-2\right)$$ but until now I still cannot solve it for any arbitrarily taken function $\vec{g}(t)$ of my choice.