# Double integral/change of variable

by hnh
Tags: change of variable, jacobian, order of integration, substitution
 P: 7 Hello, it may look long but if written down- is really quite short, I have only included a detailed explanation to try and be more clear. My question is; I am doing a change of variable which doesn't seem to 'fit' the theory of finding the jacobian to make the transformation. It seems to me that I make a straightforward substitution: The integral is \int\int_A f'(t) ln( f(x) - f(t) + t - x)dxdt In my area, x <= t so the order cannot be switched. I make the following change of variable: Let r=x-f(x) and h(r) = x and let s=t-f(t) and h(s)=t notice that this gives f(x)=h(r)-r and f(t)=h(s)-s, hence f'(t)dxdt = dxf'(t)dt = h'(r)dr(h'(s) - 1)ds=h'(r)(h'(s)-1)drds. Now the integral appears to be \int\int_A' ln(r-s)h'(r)(h'(s)-1)drds Is this correct-can I do this change of variable without actually doing the jacobian. It seems that I do not actually have the integral \int\int f(x,y) dxdt in this case to make the transformation with the jacobian so I am wondering if there is anything wrong with the way I have done this?
 P: 162 I think it would work because your r and s variables are not each one dependent on x and t at the same time, so it's like you're doing two one-variable changes, so you really are doing the Jacobian thingy even if you don't realise that you are, if you know what I mean. You have: $$r(x) \equiv x -f(x) \ \ \rightarrow \ \ x=h(r);$$ $$s(t) \equiv t-f(t) \ \ \rightarrow \ \ t=h(s);$$ $$f(x)=x-r(x) \ \ \rightarrow \ \ f(x(r)) \equiv F(r) = h(r)-r;$$ $$f(t)=t-s(t) \ \ \rightarrow \ \ f(t(s)) \equiv F(s) = h(s)-s;$$ The Jacobian would be: $$J = det \left( \begin{array}{cc} \frac {\partial x} {\partial r} & \frac {\partial t} {\partial r}\\ \frac {\partial x} {\partial s} & \frac {\partial t} {\partial s} \end{array} \right) = det \left( \begin{array}{cc} h'(r) & 0\\ 0 & h'(s) \end{array} \right) = h'(r) h'(s)$$ so you have: $$dxdt = J \ dr ds = h'(r) h'(s) dr ds$$ and $$ln|f(x(r))-f(t(s))+t(s)-x(r)| = ln|s-r|$$ and $$f'(t) \ = \ \frac {df} {dt} = \left( 1 - \frac {ds} {dt} \right)$$ And so: $$\iint_\textrm{A} f'(t) \ ln|f(x)-f(t)+t-x|dxdt = \iint_\textrm{A'} \left( 1 - \frac {ds} {dt} \right) ln|s-r|h'(r) h'(s) dr ds= \iint_\textrm{A'} \left( h'(s)- \frac {ds} {dt} \frac {dh} {ds} \right) ln|s-r|h'(r) dr ds$$ $$= \iint_\textrm{A'} \left( h'(s)- \frac {ds} {dt} \frac {dt} {ds} \right) ln|s-r|h'(r) dr ds= \iint_\textrm{A'} \left( h'(s)- 1 \right) ln|s-r|h'(r) dr ds$$ which is what you got so I think it's correct :)
 P: 7 Thank you very much-that is a huge help!
P: 162

## Double integral/change of variable

You're welcome

 Related Discussions Calculus & Beyond Homework 0 Calculus & Beyond Homework 2 Introductory Physics Homework 0 Introductory Physics Homework 0