# 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 $$\frac12 \int_{\lambda} (bz-cy)dx + (cx-az)dy + (ay-bx)dz$$

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.

• Ok, I understand that dr = (dx,dy,dz). Where I am confused is how n x r = F...? So n = (a,b,c) is normal to the plane curve. Is r = (x,y,z) just a position vector that is on the surface of the curve? I see how it all comes together but I just can't see where r x n = F comes from. – user68203 Apr 11 '13 at 6:32
• I was trying to illustrate by analogy to the simple case you already know. $\hat{n} \times \vec{F} \cdot d\vec{r}$ is coordinate independent, so if it computes an area in the $xy$ plane, it does so for any plane. – Ron Gordon Apr 11 '13 at 6:51
• where does the 1/2 come from? – user68203 Apr 12 '13 at 14:40
• Each of the area elements of the integrand represents an infinitesimal triangular element with respect to a common 3rd point in the interior of the closed curve; the area of this element is one-half of a determinant formed by these points, which turns out to be the integrand. See, for example, this: people.richland.edu/james/lecture/m116/matrices/… – Ron Gordon Apr 12 '13 at 14:45
• Ok, so it comes from taking small triangles from the surface and when you find the area you just take the 1/2 out of the integration...? – user68203 Apr 12 '13 at 14:49

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 $$S(A) = \int_A dS = \int_A curl(f).n dS = \int_{\partial A} f.dr = \frac{1}{2}\int_\lambda (bz-cy)dx + (cx-az)dy + (ay-bx) dz$$