Double integrals: when a parallelogram approximates a curvy area

In summary, the conversation discusses the topic of change of variables in double integrals and provides premises for this concept. It also introduces the idea of using parallelograms to approximate the area of a curvy region. The connection between differentiable functions and linear functions is mentioned, and the conversation ends with a reference to the proof of the Leibniz rule.
  • #1
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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  • #2
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".
 
  • #3
Hm, nobody say nothing about my question? maybe too boring?
 
  • #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?
 
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  • #5
Er... moreless. Thanks.

By the way, did you find how to interchange limits and derivatives to proof the Leibniz rule?
 
  • #6
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?
 

FAQ: Double integrals: when a parallelogram approximates a curvy area

1. What is a double integral?

A double integral is a mathematical concept used to calculate the volume under a surface in a three-dimensional space. It involves integrating a function over a two-dimensional region, often represented as a curvy area.

2. How is a double integral different from a single integral?

A single integral calculates the area under a curve in one dimension, while a double integral calculates the volume under a surface in two dimensions. In other words, a single integral is a special case of a double integral.

3. How does a parallelogram approximate a curvy area in a double integral?

In a double integral, the curvy area is divided into small rectangles or parallelograms. These parallelograms are then used to approximate the curvy area, and the sum of their areas is calculated to find the total volume under the surface.

4. What is the importance of using a parallelogram in a double integral?

Using a parallelogram in a double integral allows for a more accurate approximation of the curvy area. This is because the parallelogram can be tilted and shifted to better fit the shape of the curvy area, resulting in a more precise calculation of the volume.

5. How are double integrals used in real-life applications?

Double integrals have numerous applications in physics, engineering, and economics, among other fields. They are used to calculate the mass, center of mass, and moment of inertia of irregularly shaped objects, as well as to find the volume of 3D objects and the area of curved surfaces.

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