1
$\begingroup$

I'm trying to understand several aspects of Jacobian determinant in changing variables in multiple integral.

$[1]$ If I want to calculate double integral: $$\iint_Rf(x,y)dxdy$$ and made a substitution: $x=x(u,v)$ and $y=y(u,v)$ then using chain rule: $$dx=\frac{\partial x}{\partial u}du+\frac{\partial x}{\partial v}dv$$ $$dy=\frac{\partial y}{\partial u}du+\frac{\partial y}{\partial v}dv$$

and then: $$dxdy=\left(\frac{\partial x}{\partial u}du+\frac{\partial x}{\partial v}dv\right)\left(\frac{\partial y}{\partial u}du+\frac{\partial y}{\partial v}dv\right)$$ $$dxdy=\frac{\partial x}{\partial u}\frac{\partial y}{\partial u}du^2+\frac{\partial x}{\partial v}\frac{\partial y}{\partial v}dv^2+\left(\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}+\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}\right)dudv$$ how this last line could be rearranged to be equal the Jacobian:

$$J(u,v)=\left|\frac{\partial(x,y)}{\partial(u,v)}\right|=\left|\begin{array}{cc} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{array}\right|$$

$[2]$ Is there cases which its Jacobian $J(u,v)=0$ ? If so, what does this mean ?

$[3]$ Why when we change the variable and do the double integral we put the absolute value of Jacobian ? $$ \iint_R\,f(x(u,v),y(u,v)) \left| \frac{\partial(x,y)}{\partial(u,v)} \right|\,du\,dv$$

$\endgroup$
  • 1
    $\begingroup$ Re (1), note that the rules to compute $dxdy$ are that $dudu=dvdv=0$ and $dvdu=-dudv$. $\endgroup$ – Did Aug 8 '16 at 12:32
  • $\begingroup$ Why is that ? could you please explain ? $\endgroup$ – Mohamed Mostafa Aug 8 '16 at 12:34
  • 1
    $\begingroup$ Related, the link given in a comment there included: math.stackexchange.com/a/37156 $\endgroup$ – Did Aug 8 '16 at 12:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.