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how vector area of a closed surface is zero?

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In summary, we are discussing the concept of vector area of a closed surface and its relationship to calculus. The area of a closed surface is zero, as shown by various approaches such as using the sum of the sides of a cube or the generalized Stokes' Theorem. This concept goes beyond traditional calculus and involves other areas of mathematics such as algebra and geometry. The fundamental theorem of calculus, specifically Stokes' Theorem, is often used to explain this concept, but it is not the only approach. Overall, the area of a closed surface is a fundamental topic in mathematics with various applications and implications.

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how vector area of a closed surface is zero?

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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?Manisha Punia said:how vector area of a closed surface is zero?

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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.

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I did say thelurflurf said:Mandelbroth 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.

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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.

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

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

A closed surface is a three-dimensional shape that completely encloses a finite amount of space. It has no boundary and can be represented by a continuous vector field.

The vector area of a closed surface is calculated by integrating the cross product of the surface's unit normal vector and the differential area vector over the entire surface.

The vector area of a closed surface is always zero because the integral of the cross product of the unit normal vector and the differential area vector is equal to zero. This is due to the fact that the surface has no boundary, so the contributions from each point on the surface cancel each other out.

No, the vector area of a closed surface can never be non-zero. This is a fundamental property of closed surfaces in vector calculus.

The vector area of a closed surface being zero means that there is no net flux of the vector field across the surface. This has important applications in physics, such as in the study of electric and magnetic fields.

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