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Double integrals: when a parallelogram approximates a curvy area

  1. Oct 19, 2005 #1
    Sirs, I request your assistance. I am reading chapter 13.9 of James Stewart Calculus (change of variables in double integrals).

    Premises.- Let (x,y) = T(u,v). The function T has inverse, both are continuously differentiable, etc.

    Let’s take a very little rectangle S in the uv plane. The point left and down we will denote (u0, v0).

    1. Any point of the low base of S is (u, v0).

    We apply the transformation T to rectangle S and obtain a curvy area which we will denote R.

    Now, let “p” be a vectorial function of real variable such that:
    p(u) = ( x(u, v0), y(u, v0) ).
    Then (this I understand) p’(u0) = i x_u(u0,v0) + j y_u(u0,v0). This result is "A".

    (x_u is the derivative of x with respect to u).

    2. Any point of the left side of rectangle S is (u0, v).

    Let “g” be a vectorial function of real variable such that:
    g(v) = ( x(u0, v), y(u0, v) ).
    Then (this I understand) g’(v0) = i x_v(u0,v0) + j y_v(u0,v0). This result is "B".

    Now comes the part I dont understand. Stewart says: “With results A and B we can approximate the area of R by way of the parallelogram defined by these two vectors:
    - delta “u” p’(u0) and
    - delta “v” g’(v0).

    Why is this? I know that it has something to do with the fact that differentiable functions can be approximate by linear functions, but I fail to see the conexión with the boundary of the curvy area R.

    Please help.
     
  2. jcsd
  3. Oct 19, 2005 #2
    I forgot to state that the lenghts of sides of little rectangle S are delta "u" and delta "v".
     
  4. Oct 20, 2005 #3
    Hm, nobody say nothing about my question? maybe too boring?
     
  5. Oct 20, 2005 #4
    I don't have the book, and I'm just going from the title since the post was kind of difficult to follow.

    By rectangles approximating a curvy area, do you mean surface area? What is found is two tangent vectors, call them ru and rv. If you know anything about vectors, then you know that [itex]|\Delta u \vec{r}_u\times \Delta v \vec{r}_v|[/itex] is the area of the parallelogram formed by these vectors. This parallelogram is tangent to the surface f(x,y) at one point. Letting each parallelogram be very small and taking the sum (over the xy plane) we have the following:

    [tex]\text{SA}=\iint\limits_D\left|\vec{r}_u\times\vec{r}_v\right|\,du\,dv[/tex]

    Does this make any sense?
     
    Last edited: Oct 20, 2005
  6. Oct 20, 2005 #5
    Er... moreless. Thanks.

    By the way, did you find how to interchange limits and derivatives to proof the Leibniz rule?
     
  7. Oct 20, 2005 #6
    Yes I did. Were you following that post as well?
     
  8. Oct 20, 2005 #7
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