Proving Change of Variables Formula for Double Integral w/ Chain Rule

[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?

 

[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?

 

[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.

 

[matt grime]

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

 

[HallsofIvy]

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?

?? 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.

 

[Benny]

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