Double integral/change of variableby hnh Tags: change of variable, jacobian, order of integration, substitution 

#1
Mar1109, 10:06 AM

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=xf(x) and h(r) = x and let s=tf(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(rs)h'(r)(h'(s)1)drds Is this correctcan 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? 



#2
Mar1209, 03:07 PM

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 onevariable 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:
[tex] r(x) \equiv x f(x) \ \ \rightarrow \ \ x=h(r); [/tex] [tex] s(t) \equiv tf(t) \ \ \rightarrow \ \ t=h(s); [/tex] [tex] f(x)=xr(x) \ \ \rightarrow \ \ f(x(r)) \equiv F(r) = h(r)r; [/tex] [tex] f(t)=ts(t) \ \ \rightarrow \ \ f(t(s)) \equiv F(s) = h(s)s; [/tex] The Jacobian would be: [tex] 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) [/tex] so you have: [tex] dxdt = J \ dr ds = h'(r) h'(s) dr ds [/tex] and [tex] lnf(x(r))f(t(s))+t(s)x(r) = lnsr [/tex] and [tex] f'(t) \ = \ \frac {df} {dt} = \left( 1  \frac {ds} {dt} \right) [/tex] And so: [tex] \iint_\textrm{A} f'(t) \ lnf(x)f(t)+txdxdt = \iint_\textrm{A'} \left( 1  \frac {ds} {dt} \right) lnsrh'(r) h'(s) dr ds= \iint_\textrm{A'} \left( h'(s) \frac {ds} {dt} \frac {dh} {ds} \right) lnsrh'(r) dr ds [/tex] [tex]= \iint_\textrm{A'} \left( h'(s) \frac {ds} {dt} \frac {dt} {ds} \right) lnsrh'(r) dr ds= \iint_\textrm{A'} \left( h'(s) 1 \right) lnsrh'(r) dr ds [/tex] which is what you got so I think it's correct :) 



#3
Mar1309, 11:13 AM

P: 7

Thank you very muchthat is a huge help!




#4
Mar1409, 01:56 PM

P: 162

Double integral/change of variable
You're welcome



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