# How vector area of a closed surface is zero?

1. Oct 12, 2013

### Manisha Punia

how vector area of a closed surface is zero?

2. Oct 12, 2013

### Mandelbroth

You should explain the context, as well as your mathematical experience. There are varying ways we can answer your question. The best way to answer your question, if I'm understanding it correctly, goes beyond multivariate calculus. What course is this for?

3. Oct 12, 2013

### lurflurf

Mandelbroth can you explain what you mean by goes goes beyond calculus? That is very much a first year calculus question. It involves other mathematics (ie algebra and geometry) and is difficult to answer in general or for difficult specific examples. That is true throughout calculus. It would be helpful if
Manisha Punia made the question more specific.

Consider the (vector) area of a cube
the sides have area
i
-i
j
-j
k
-k
The area of the cube is the sum of the sides
area=i-i+j-j+k-k=0

Consider the cube with one face removed
area=-i+j-j+k-k=-i

Notice the area of the cube with one face removed is the additive inverse of the area of the removed face.

Call two surfaces whose union is closed complementary.
We are setting up a system of area with desirable properties.
-The area of the union of two surfaces is the sum of the area.
-The areas of complementary surfaces are additive inverses.
-The area of the empty surface is zero

The empty surface is complementary to any closed surface, so any closed surface has area 0.

Another approach is to assign (vector) area to the boundary of a surface. Closed surfaces have no boundary so the boundary has area 0 so the surface has area 0.

4. Oct 12, 2013

### Mandelbroth

I did say the best way, right? The best way, in my opinion, is via the generalized Stokes' Theorem.

Then again, I tend to lean on that theorem a lot more than I need to.

5. Oct 12, 2013

### lurflurf

Stokes' Theorem is a fundamental theorem of calculus. Some might say Stokes' Theorem is the fundamental theorem of calculus. Granted as I said above it involves other mathematics (ie algebra and geometry) and is difficult to show in general or for difficult specific examples. Many calculus books have Stokes' theorem towards the end and/or do little with it. A big problem is that calculus is a hodge podge of random techniques and not a unified subject.

6. Oct 14, 2013

### LCKurtz

How about $\vec i \cdot \iint_S d\vec S=\iint_S\hat i\cdot d\vec S =\iiint_V \nabla \cdot \vec i~dV = 0$ and similarly for $\vec j$ and $\vec k$, so all three components are zero.

7. Oct 15, 2013

### Mandelbroth

Not the Stokes' Thoerem I'm talking about. :tongue:

LCKurtz's method is also good (though it actually follows from the generalized Stokes' Theorem! ).