# Double integral/change of variable

 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 :)