Surface element -- why parallelograms?

  • #1

Main Question or Discussion Point

we say the surface element of x(u,v),y(u,v),z(u,v) is absolute value of the cross product
rtial%20r%7D%7B%5Cpartial%20u%7D%5Ctimes%20%5Cfrac%7B%5Cpartial%20r%7D%7B%5Cpartial%20v%7D%3E%7C.gif

or
%5Cpartial%20y%7D%7B%5Cpartial%20v%7D%2C%5Cfrac%7B%5Cpartial%20z%7D%7B%5Cpartial%20v%7D%3E%20%7C.gif

i dont understand why this is. i know that the absolute value of the cross product
gif.gif
is the area of a parallelogram formed by a,b,a+b, but i get lost after that
im guessing something happens where dudv cancels out with ∂u∂v then just leaves ∂r, but i cant really figure what it means.

anyone whos got into computer graphics can tell you triangles are the building blocks of surfaces, because any 3 points occupy the same plane. why a parallelogram then?
 
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Answers and Replies

  • #2
Simon Bridge
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Because that is how the maths is simplest.
It makes no difference otherwise since the main difference between above and the computer approximation to a continuous surface is that the parallelogram is arbitrarily small.

Note: the parallelogram is not the only way of viewing the cross product.
 
  • #3
Stephen Tashi
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i know that the absolute value of the cross product
gif.gif
is the area of a parallelogram formed by a,b,a+b, but i get lost after that
I'll reinterpret your question.

We could define a surface integral in a sophisticated way by saying it is the limit of processes that cover the surface with small shapes tangential to its surface and then form a sum of the area of each shape times the value of the function at the point of tangency.

We can define a surface integral in a less sophisticated way by specifying each small shape exactly. For the surface [itex] (x(u,v),y(u,v),z(u,v)) [/itex] we can think of [itex] || <\frac{\partial x}{\partial u}, \frac{\partial y}{\partial u}, \frac{\partial z}{\partial u}> \times <\frac{\partial x}{\partial v}, \frac{\partial y}{\partial v}, \frac{\partial z}{\partial v} >|| du dv [/itex] as

[itex] || du <\frac{\partial x}{\partial u}, \frac{\partial y}{\partial u}, \frac{\partial z}{\partial u}> \times dv <\frac{\partial x}{\partial v}, \frac{\partial y}{\partial v}, \frac{\partial z}{\partial v} >|| [/itex]

In other words, think of [itex] du [/itex] and [itex] dv [/itex] as scalars that multiply the vectors involved so the expression is the area of a small parallelogram.

The two gradient vectors conveniently define a plane tangent to the surface. Multiplying the gradient vectors by small scalars when we take the cross product gives us the area of a small parallelogram lying in the tangent plane. This still leaves us open the question of why we can use the gradient vectors without normalizing them each to have length 1. Without that normalization we are using bigger parallelograms where the shape of the surface changes rapidly. It seems we should use smaller parallelograms there instead.
 
  • #4
HallsofIvy
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If you start with a rectangle ("[itex]\Delta x[/itex]" by "[itex]\Delta y[/itex]) in the xy-plane and project onto a plane that is NOT parallel to the xy-plane, the projection is a parallelogram. For example, if the rectangle with vertices at (0, 0), (1, 0), (1, 1), and (0, 1) is projected to the plane z= a+ bx+ cy, the vertices are projected to (0, 0, a), (1, 0, a+ b), (1, 1, a+ b+ c), and (0, 1, a+ c).

It is easy to show that opposite sides of that quadrilateral, such as (0, 0, a) to (1, 0, a+ b) and (0, 1, a+ c) to (1, 1, a+ b+ c) have the same length: [itex]\sqrt{(0- 1)^2+ (0- 0)^2+ (a- (a+ b))^2}= \sqrt{1+ b^2}[/itex] and [itex]\sqrt{(0- 1)^2+ (1- 1)^2+ ((a+ c)- (a+ b+ c)^2}= \sqrt{1+ b^2}[/itex].

But the angles are NOT right angles: the line from (0, 0, a) to (1, 0, a+ b) can be written in parametric equations as x= t, y= 0, z= a+ bt, with t from 0 to 1. A vector in the direction of that line is <1, 0, b>. The line from (0, 0, a) to (0, 1, a+ c) can be written in parametric equations as x= 0, y= t, z= a+ ct with t from 0 to 1. A vector in the direction of that line is <0, 1, c>. The dot product of those two vectors is 0(1)+ 1(0)+ bc= bc, NOT 0.

A quadrilateral with opposite sided of the same length, but angle not right angles, is a general parallelogram.
 

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