Matrices, Scalars, Vectors and Vector Calculus 3

After introducing some mathematical machinery with our first and second posts it is now time for us to look into some Newtonian Physics, after a brief look into vector integration.

— 1. Vector Integration —

When dealing with vectors and the mathematical operation we have three basic options:

  • Volume integration
  • Surface integration
  • Line (contour) integration

The result of integrating a vector, {\vec{A}=\vec{A}(x_i)}, over a volume is also a vector and the result is given by the following expression:

\displaystyle \int_V \vec{A}dv = \left( \int_V A_1 dv, \int_V A_2 dv, \int_V A_3 dv \right) \ \ \ \ \ (1)

 

Hence, the result of vector integration is just three separate integration operations (one of each spatial dimension).

The result of integrating the projection of a vector {\vec{A}=\vec{A}(x_1)} over an area is what is called surface integration.

Surface integration is always done with the normal component of {\vec{A}} over the surface {S} in question. Thus what we need to define first is the normal of a surface at a given point. {d\vec{a}=\vec{n}da} will be this normal. We still have the ambiguity of having two possible directions for the normal at any given point, but this is taken care of by defining the normal to be on the outward direction of a closed surface.

Hence the quantity of interest is {\vec{A}\cdot d\vec{a}=\vec{A}\cdot \vec{n} da} ({da_1=dx_2dx_3} for example) with

\displaystyle \int_S \vec{A}\cdot d\vec{a} = \int_S \vec{A}\cdot \vec{n}da \ \ \ \ \ (2)

 

As for the line integral it is define along the path between two points {B} and {C}. Again we have to consider the normal of {\vec{A}=\vec{A}(x_i)}, but this time the quantity of interest is:

\displaystyle \int_{BC} \vec{A}\cdot d\vec{s} \ \ \ \ \ (3)

 

The quantity {d\vec{s}} is an element of length along {BC} and is taken to be positive along the direction in which the path is defined.

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14 comments on “Matrices, Scalars, Vectors and Vector Calculus 3

  1. joeschmo26 says:

    I believe I am onboard until this point.

  2. joeschmo26 says:

    Also,are there exercise’s we are supposed to attempt? If there are, I am not seeing them in the body of the blog. (Note: I have never used Surface Integrals to solve any physical problems, but I am aware that they are used when solving problems involving electric and magnetic fields. I guess they are crucial to the study of any field in general, and suppose “why” that is the case will be explained further down the road?)

  3. joeschmo26 says:

    {\vec{A}=\vec{A}(x_i)}

  4. joeschmo26 says:

    $atex{\vec{A}=\vec{A}(x_i)}$

  5. joeschmo26 says:

    I guess I don’t get this LateX bit after all. However, I have done some research on application of the surface integral, and have a better handle on it…So,not all is lost.

  6. ateixeira says:

    Joe your second attempt was almost right. You have to right $latex, then leave a space write your mathematical expression, leave a space and finish with $. Give it a try

  7. joeschmo26 says:

    Ok, thanks for the tip.

    {\vec{A}=\vec{A}(x_i)}

  8. joeschmo26 says:

    Well, I can see that this will take some practice, and many examples of equations so I learn the codes for the various operators.

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