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how vector area of a closed surface is zero?

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- Thread starter Manisha Punia
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- #1

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how vector area of a closed surface is zero?

- #2

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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?how vector area of a closed surface is zero?

- #3

lurflurf

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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

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I did say theMandelbroth can you explain what you mean by goes goes beyond calculus? That is very much a first year calculus question.

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

- #5

lurflurf

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- #6

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=\iiint_V \nabla \cdot \vec i~dV = 0## and similarly for ##\vec j## and ##\vec k##, so all three components are zero.

- #7

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

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

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