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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Double integrals: when a parallelogram approximates a curvy area

**Physics Forums | Science Articles, Homework Help, Discussion**