Sirs, I request your assistance. I am reading chapter 13.9 of James Stewart Calculus (change of variables in double integrals).(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Homework Help: Double integrals: when a parallelogram approximates a curvy area

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