Using Stoke's Theorem to determine plane area enclosed by simple closed plane curve

Homework Statement

Let C be a simple closed plane curve in space. Let n = ai+bj+ck be a unit vector normal to the plane of C and let the direction on C match that of n. Prove that

(1/2)∫[(bz-cy)dx+(cx-az)dy+(ay-bx)dz]

equals the plane area enclosed by C.

What does the integral reduce to when C is in the xy-plane?

Stoke's Theorem

F.ds=∫(∇×F).dS

The Attempt at a Solution

I really have no idea where to start. Any help would be much appreciated.

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LCKurtz
Homework Helper
Gold Member

Homework Statement

Let C be a simple closed plane curve in space. Let n = ai+bj+ck be a unit vector normal to the plane of C and let the direction on C match that of n. Prove that

(1/2)∫[(bz-cy)dx+(cx-az)dy+(ay-bx)dz]

equals the plane area enclosed by C.

What does the integral reduce to when C is in the xy-plane?

Stoke's Theorem

F.ds=∫(∇×F).dS

The Attempt at a Solution

I really have no idea where to start. Any help would be much appreciated.
Why don't you start by calculating ##\nabla \times \vec F## and plug it in the right side of Stokes' Theorem?

Why don't you start by calculating ##\nabla \times \vec F## and plug it in the right side of Stokes' Theorem?
I don't know what F is

LCKurtz
Homework Helper
Gold Member

You are given a line integral, written in the form ##\oint \vec F\cdot d\vec R##. Can't you pick ##\vec F## out of that?

You are given a line integral, written in the form ##\oint \vec F\cdot d\vec R##. Can't you pick ##\vec F## out of that?
I think I might have been reading the question incorrectly (well hopefully). I feel like an idiot. I'll give it a go and hopefully it works out. Thanks LCKurtz

Got it out

Thanks again for your help LCKurtz