Changing one variable in a multiple integral
$begingroup$
Lets suppose I want to calculate
$$I=int_{mathbb{R}^ntimesmathbb{R}^m} f(x,y):mathrm{d}x:mathrm{d}y.$$
If I make $u=h(x,y)$ and $v=y$, will the change in variables be
$$I=int_{mathbb{R}^ntimesmathbb{R}^m} f(g^{-1}(u),v)|det J_{g^{-1}} |:mathrm{d}u:mathrm{d}v,$$
where $g$ is the function $xmapsto h(x,v)$? If so, why?
real-analysis multivariable-calculus multiple-integral
$endgroup$
add a comment |
$begingroup$
Lets suppose I want to calculate
$$I=int_{mathbb{R}^ntimesmathbb{R}^m} f(x,y):mathrm{d}x:mathrm{d}y.$$
If I make $u=h(x,y)$ and $v=y$, will the change in variables be
$$I=int_{mathbb{R}^ntimesmathbb{R}^m} f(g^{-1}(u),v)|det J_{g^{-1}} |:mathrm{d}u:mathrm{d}v,$$
where $g$ is the function $xmapsto h(x,v)$? If so, why?
real-analysis multivariable-calculus multiple-integral
$endgroup$
add a comment |
$begingroup$
Lets suppose I want to calculate
$$I=int_{mathbb{R}^ntimesmathbb{R}^m} f(x,y):mathrm{d}x:mathrm{d}y.$$
If I make $u=h(x,y)$ and $v=y$, will the change in variables be
$$I=int_{mathbb{R}^ntimesmathbb{R}^m} f(g^{-1}(u),v)|det J_{g^{-1}} |:mathrm{d}u:mathrm{d}v,$$
where $g$ is the function $xmapsto h(x,v)$? If so, why?
real-analysis multivariable-calculus multiple-integral
$endgroup$
Lets suppose I want to calculate
$$I=int_{mathbb{R}^ntimesmathbb{R}^m} f(x,y):mathrm{d}x:mathrm{d}y.$$
If I make $u=h(x,y)$ and $v=y$, will the change in variables be
$$I=int_{mathbb{R}^ntimesmathbb{R}^m} f(g^{-1}(u),v)|det J_{g^{-1}} |:mathrm{d}u:mathrm{d}v,$$
where $g$ is the function $xmapsto h(x,v)$? If so, why?
real-analysis multivariable-calculus multiple-integral
real-analysis multivariable-calculus multiple-integral
edited Jan 5 at 8:49
Gabriel Ribeiro
asked Jan 5 at 8:06
Gabriel RibeiroGabriel Ribeiro
1,454523
1,454523
add a comment |
add a comment |
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