tough question in Line integral / Multivariable Calculas

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Let $C$ be a curve in the $(x − y)$-plane. For every point $(x, y)$ of $C$ let $u(x, y)$ denote the unit vector in the direction of the tangent line to $C$ at $(x, y)$. Let $S$ be the surface obtained by taking the union of all straight line segments connecting $(1, 2, 3)$ to points of $C$. Express the area of $S$ as an integral of the first type, on the curve $C$, of some function of $x$ and $y$. (hint: try to use the function $u(x, y)$.)

really Hard question , i couldn't understand how to use the fact that line integral will help here since i don't have a function $f(x,y)$ to calculate $ \int_{C} f(x(t),y(t))\sqrt{x'(t)^2 + y'(t)^2 }\ dt$ also i can parameterize $S$ like that : $S=:k(1-x(t),2-y(t),3) , k\in[0,1]$ where $(x(t),y(t),0)$ is the curve $C$

and why $u(x,y)$ is given here. Unit vector

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1
On BEST ANSWER

$c(t) = (x(t),y(t), 0)\\ \frac {dc}{dt} = u(t)$

The parmeterization of $S$ should be

$S = (k +(1-k) x,2 k + (1-k) y, 3k)$

or

$((1-k) +k x,2(1- k) + k) y(t), 3(1-k))$

with

$0\le k \le 1$

$dS = \|\frac {\partial S}{dk} \times \frac {\partial S}{dt}\|$

$\frac {\partial S}{dk} = (1-x, 2-y,3) = (1,2,3) - c(t)$

$\frac {\partial S}{dt} = (1-k)u(t)$

$\int_0^1 (1-k) dk \int_0^t \|(1,2,3)\times u(t) - c(t)\times u(t)\| \ dt$

$\frac 12 \int_0^t \|(1,2,3)\times u(t) - c(t)\times u(t)\| \ dt$

0
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This is a very non-rigorous derivation.

Consider a point $C(t) = (x(t),y(t),0) \in C$ and a second point infinitesimally close to it: $$C(t+dt) = (x(t+dt),y(t+dt),0) = (x(t)+\dot{x}(t)\,dt,y(t)+\dot{y}(t)\,dt,0)$$ We need to calculate the infinitesimal area $dA$ of the triangle with vertices $C(t),C(t+dt)$ and $(1,2,3)$.

The area is given by \begin{align} dA &= \frac12 \|(C(t+dt) - C(t)) \times ((1,2,3) - C(t))\| \\ &= \frac12 \|(\dot{x}\,dt,\dot{y}\,dt,0) \times (1-x,2-y,3)\|\\ &= \frac{dt}2 \|(\dot{x},\dot{y},0) \times (1-x(t),2-y(t),3)\|\\ &= \frac{dt}2 \|(3\dot{y},-3\dot{x},(2-y)\dot{x}+(x-1)\dot{y})\|\\ \end{align}

so $$A = \int_C\,dA = \frac12\int_{t}\|(3\dot{y}(t),-3\dot{x}(t),(2-y(t))\dot{x}(t)+(x(t)-1)\dot{y}(t))\|\,dt$$