Find area of a simple, smooth, closed curve lying in a plane I was given this question in class and I assume it is a spin off of Green's theorem for finding the area of a closed curve $\lambda$ in 2D but expanded to 3D I believe. Anyways I am pretty confused about it so if anyone could help I would appreciate it,.
Question: Let $\lambda$ be a simple closed smooth space curve that lies in a plane with
unit normal vector n = (a, b, c) and has positive orientation with re-
spect to the normal vector n of the plane. Show that the plane area
enclosed by $\lambda$ is $$\frac12 \int_{\lambda} (bz-cy)dx + (cx-az)dy + (ay-bx)dz$$
 A: The Green's theorem we know takes the form for the area enclosed by a closed curve:
$$A = \iint_A dx \, dy = \frac{1}{2} \oint_C (-y\, dx + x\, dy) = \frac{1}{2} \oint_C \vec{F}\cdot d\vec{r}$$
where $\vec{F}=(-y,x,0)$ and $d\vec{r} = (dx,dy,dz)$.  Note that this is for the $xy$ plane, in which the normal vector is $\hat{n}=(0,0,1)$.  You can see that, for a position vector $\vec{r}=(x,y,z)$:
$$\hat{n} \times \vec{r} = \vec{F}$$
This is true in general.  For an arbitrary plane in which $\hat{n}=(a,b,c)$,
$$\vec{F}=\hat{n} \times \vec{r} = (b z-c y,c x-a z,a y-b x)$$
Therefore
$$A =  \frac{1}{2} \oint_C \vec{F}\cdot d\vec{r} = \frac{1}{2} \oint_C [(b z-c y) dx + (c x-a z) dy + (a y-b x) dz]$$
as was to be shown.
A: This is immidiate using the stoke's theorem where $S$ being surface measure and $A$ be surface enclosed by the curve, you take
$$ f(x,y,z) = (1/2)(bz-cy,cx-az,ay-bx) $$ And you find $ curl(f) = (a,b,c) = n $. and you have $n.n = 1 $ and $\partial A = \lambda $. so from stoke's theorem you get area of the surface S(A) as 
$$ S(A) = \int_A dS = \int_A curl(f).n dS = \int_{\partial A} f.dr 
 = \frac{1}{2}\int_\lambda (bz-cy)dx + (cx-az)dy + (ay-bx) dz $$
