# Chain rule

1. Nov 26, 2005

### Benny

Hi, I'm having trouble understanding the solution to a question from my book. I think it's got something to do with the chain rule. The problem is to prove the change of variables formula for a double integral for the case f(x,y) = 1 using Green's theorem.

$$\int\limits_{}^{} {\int\limits_R^{} {dxdy} } = \int\limits_{}^{} {\int\limits_S^{} {\left| {\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}} \right|} } dudv$$

The solution starts off by using some equation in the theory section allows for the following equality to be established.

$$\int\limits_{}^{} {\int\limits_R^{} {dxdy} } = A\left( R \right) = \int\limits_{\partial R}^{} {xdy}$$

Then it says x = g(u,v) and $$dy = \frac{{\partial h}}{{\partial u}}du + \frac{{\partial h}}{{\partial v}}dv$$.

I don't understand how the expression for dy is arrived at. The variable y is a function of u and v. I can't see a way to use the chain rule here. Can someone please explain how the dy part is arrived at?

Last edited: Nov 26, 2005
2. Nov 26, 2005

### EnumaElish

I am not a mathematician, but you have the integrand as xdy, and y is apparently equal to h(u,v) so dy = h_u du + h_v dv. Does that answer?

3. Nov 28, 2005

### Benny

It's just that I can't see how they got the expression for dy. I'm wondering if it is something similar to for example when you have y = x^2 then from that you get dy = 2x dx.

4. Nov 28, 2005

### matt grime

That's just how it (the 'chain rule') works for d's, it's obvious, isn't it?

5. Nov 28, 2005

### HallsofIvy

?? That is the chain rule, in differential form. Perhaps you are more familiar with the chain rule as
$$\frac{dy}{dx} = \frac{{\partial h}}{{\partial u}}\frac{du}{dx} + \frac{{\partial h}}{{\partial v}}\frac{dv}{dx}$$
From that, it follows from the definition of differentials that
$$dy= \frac{dy}{dx}dx$$
so that
$$dy= \frac{dy}{dx}dx = \frac{{\partial h}}{{\partial u}}\frac{du}{dx}dx + \frac{{\partial h}}{{\partial v}}\frac{dv}{dx}dx$$
$$dy = \frac{{\partial h}}{{\partial u}}du + \frac{{\partial h}}{{\partial v}}dv$$

Notice that I have assumed that u and v are functions of x. The nice thing about differential form is that you don't have to assume any specific parameter.

6. Nov 28, 2005

### Benny

Ok I see how it's arrived at now. Thanks for the help.