Double integrals: when a parallelogram approximates a curvy area

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Homework Help Overview

The discussion revolves around the application of double integrals and the transformation of coordinates in calculus, specifically focusing on approximating a curvy area defined by a transformation from the uv-plane to the xy-plane using parallelograms. The original poster seeks clarification on the connection between the tangent vectors derived from the transformation and the approximation of the area of the curvy region.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand why the area of the curvy region can be approximated by the parallelogram formed by the tangent vectors derived from the transformation. They express uncertainty about the relationship between differentiable functions and linear approximations in this context.
  • Another participant introduces the concept of tangent vectors and relates them to the area of the parallelogram, suggesting a connection to surface area and the use of cross products in vector calculus.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the geometric interpretation of the tangent vectors and their role in approximating the area, but no consensus has been reached on the original poster's specific question.

Contextual Notes

The original poster notes the dimensions of the rectangle in the uv-plane as delta "u" and delta "v", which may be relevant to the approximation being discussed. There is also mention of a separate inquiry regarding the interchange of limits and derivatives related to the Leibniz rule, indicating a broader context of calculus concepts being explored.

Castilla
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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 don't 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.
 
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Castilla said:
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 don't 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.

I forgot to state that the lenghts of sides of little rectangle S are delta "u" and delta "v".
 
Hm, nobody say nothing about my question? maybe too boring?
 
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:
Er... moreless. Thanks.

By the way, did you find how to interchange limits and derivatives to proof the Leibniz rule?
 
Castilla said:
Er... moreless. Thanks.
By the way, did you find how to interchange limits and derivatives to proof the Leibniz rule?
Yes I did. Were you following that post as well?
 

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