Calculate region using jacobian determinant of substitution












0












$begingroup$


Problem:



Let $B$ be the region in the first quadrant of $mathbb R^2$ restricted by the curves: $xy=1, xy=3, x^2-y^2=1, x^2-y^2=4$.



Calculate $int_B(x^2+y^2)dxdy$.



Hint: Substitute $u=xy$ and $v=x^2-y^2$



Solution:



The jacobian-determinant of the map $begin{pmatrix}u\vend{pmatrix}(x,y)$ is



$detbegin{pmatrix}y&x\2x&-2yend{pmatrix}=-2(x^2+y^2)$



So we get



$int_B(x^2+y^2)dxdy=frac{1}{2}int_1^3duint_1^4dv=3$



Question:
I think I do get the basic principle and whats going on. What I don't get is the very first sentence of the solution. They say they take the jacobian-determinant of the map $begin{pmatrix}u\vend{pmatrix}(x,y)$.



Now for me, that's: $begin{pmatrix}u\vend{pmatrix}(x,y)=begin{pmatrix}ux & uv \ vx & vyend{pmatrix}$
but that confuses me. Is that the correct intepretation of $begin{pmatrix}u\vend{pmatrix}(x,y)$? nd if so, how do we get that?










share|cite|improve this question









$endgroup$












  • $begingroup$
    Simply it says that the map $(u,v)^T$ is a function of the variables $x$ and $y$. Its like putting $vec{x}=(x,y)^T=vec{r}(x,y)$. It is not a Kronecker product.
    $endgroup$
    – HBR
    Jan 7 at 15:00






  • 1




    $begingroup$
    The real mapping happens via the Jacobian transformation matrix $J$ from the $(x,y)$ coordinate system to the $(u,v)$ system.
    $endgroup$
    – WarreG
    Jan 7 at 15:20










  • $begingroup$
    ah... (x,y) are arguments here! Now that makes way more sense. Thanks. You should make an answer for that.
    $endgroup$
    – xotix
    Jan 8 at 21:49
















0












$begingroup$


Problem:



Let $B$ be the region in the first quadrant of $mathbb R^2$ restricted by the curves: $xy=1, xy=3, x^2-y^2=1, x^2-y^2=4$.



Calculate $int_B(x^2+y^2)dxdy$.



Hint: Substitute $u=xy$ and $v=x^2-y^2$



Solution:



The jacobian-determinant of the map $begin{pmatrix}u\vend{pmatrix}(x,y)$ is



$detbegin{pmatrix}y&x\2x&-2yend{pmatrix}=-2(x^2+y^2)$



So we get



$int_B(x^2+y^2)dxdy=frac{1}{2}int_1^3duint_1^4dv=3$



Question:
I think I do get the basic principle and whats going on. What I don't get is the very first sentence of the solution. They say they take the jacobian-determinant of the map $begin{pmatrix}u\vend{pmatrix}(x,y)$.



Now for me, that's: $begin{pmatrix}u\vend{pmatrix}(x,y)=begin{pmatrix}ux & uv \ vx & vyend{pmatrix}$
but that confuses me. Is that the correct intepretation of $begin{pmatrix}u\vend{pmatrix}(x,y)$? nd if so, how do we get that?










share|cite|improve this question









$endgroup$












  • $begingroup$
    Simply it says that the map $(u,v)^T$ is a function of the variables $x$ and $y$. Its like putting $vec{x}=(x,y)^T=vec{r}(x,y)$. It is not a Kronecker product.
    $endgroup$
    – HBR
    Jan 7 at 15:00






  • 1




    $begingroup$
    The real mapping happens via the Jacobian transformation matrix $J$ from the $(x,y)$ coordinate system to the $(u,v)$ system.
    $endgroup$
    – WarreG
    Jan 7 at 15:20










  • $begingroup$
    ah... (x,y) are arguments here! Now that makes way more sense. Thanks. You should make an answer for that.
    $endgroup$
    – xotix
    Jan 8 at 21:49














0












0








0





$begingroup$


Problem:



Let $B$ be the region in the first quadrant of $mathbb R^2$ restricted by the curves: $xy=1, xy=3, x^2-y^2=1, x^2-y^2=4$.



Calculate $int_B(x^2+y^2)dxdy$.



Hint: Substitute $u=xy$ and $v=x^2-y^2$



Solution:



The jacobian-determinant of the map $begin{pmatrix}u\vend{pmatrix}(x,y)$ is



$detbegin{pmatrix}y&x\2x&-2yend{pmatrix}=-2(x^2+y^2)$



So we get



$int_B(x^2+y^2)dxdy=frac{1}{2}int_1^3duint_1^4dv=3$



Question:
I think I do get the basic principle and whats going on. What I don't get is the very first sentence of the solution. They say they take the jacobian-determinant of the map $begin{pmatrix}u\vend{pmatrix}(x,y)$.



Now for me, that's: $begin{pmatrix}u\vend{pmatrix}(x,y)=begin{pmatrix}ux & uv \ vx & vyend{pmatrix}$
but that confuses me. Is that the correct intepretation of $begin{pmatrix}u\vend{pmatrix}(x,y)$? nd if so, how do we get that?










share|cite|improve this question









$endgroup$




Problem:



Let $B$ be the region in the first quadrant of $mathbb R^2$ restricted by the curves: $xy=1, xy=3, x^2-y^2=1, x^2-y^2=4$.



Calculate $int_B(x^2+y^2)dxdy$.



Hint: Substitute $u=xy$ and $v=x^2-y^2$



Solution:



The jacobian-determinant of the map $begin{pmatrix}u\vend{pmatrix}(x,y)$ is



$detbegin{pmatrix}y&x\2x&-2yend{pmatrix}=-2(x^2+y^2)$



So we get



$int_B(x^2+y^2)dxdy=frac{1}{2}int_1^3duint_1^4dv=3$



Question:
I think I do get the basic principle and whats going on. What I don't get is the very first sentence of the solution. They say they take the jacobian-determinant of the map $begin{pmatrix}u\vend{pmatrix}(x,y)$.



Now for me, that's: $begin{pmatrix}u\vend{pmatrix}(x,y)=begin{pmatrix}ux & uv \ vx & vyend{pmatrix}$
but that confuses me. Is that the correct intepretation of $begin{pmatrix}u\vend{pmatrix}(x,y)$? nd if so, how do we get that?







calculus jacobian






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share|cite|improve this question











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asked Jan 7 at 14:55









xotixxotix

291411




291411












  • $begingroup$
    Simply it says that the map $(u,v)^T$ is a function of the variables $x$ and $y$. Its like putting $vec{x}=(x,y)^T=vec{r}(x,y)$. It is not a Kronecker product.
    $endgroup$
    – HBR
    Jan 7 at 15:00






  • 1




    $begingroup$
    The real mapping happens via the Jacobian transformation matrix $J$ from the $(x,y)$ coordinate system to the $(u,v)$ system.
    $endgroup$
    – WarreG
    Jan 7 at 15:20










  • $begingroup$
    ah... (x,y) are arguments here! Now that makes way more sense. Thanks. You should make an answer for that.
    $endgroup$
    – xotix
    Jan 8 at 21:49


















  • $begingroup$
    Simply it says that the map $(u,v)^T$ is a function of the variables $x$ and $y$. Its like putting $vec{x}=(x,y)^T=vec{r}(x,y)$. It is not a Kronecker product.
    $endgroup$
    – HBR
    Jan 7 at 15:00






  • 1




    $begingroup$
    The real mapping happens via the Jacobian transformation matrix $J$ from the $(x,y)$ coordinate system to the $(u,v)$ system.
    $endgroup$
    – WarreG
    Jan 7 at 15:20










  • $begingroup$
    ah... (x,y) are arguments here! Now that makes way more sense. Thanks. You should make an answer for that.
    $endgroup$
    – xotix
    Jan 8 at 21:49
















$begingroup$
Simply it says that the map $(u,v)^T$ is a function of the variables $x$ and $y$. Its like putting $vec{x}=(x,y)^T=vec{r}(x,y)$. It is not a Kronecker product.
$endgroup$
– HBR
Jan 7 at 15:00




$begingroup$
Simply it says that the map $(u,v)^T$ is a function of the variables $x$ and $y$. Its like putting $vec{x}=(x,y)^T=vec{r}(x,y)$. It is not a Kronecker product.
$endgroup$
– HBR
Jan 7 at 15:00




1




1




$begingroup$
The real mapping happens via the Jacobian transformation matrix $J$ from the $(x,y)$ coordinate system to the $(u,v)$ system.
$endgroup$
– WarreG
Jan 7 at 15:20




$begingroup$
The real mapping happens via the Jacobian transformation matrix $J$ from the $(x,y)$ coordinate system to the $(u,v)$ system.
$endgroup$
– WarreG
Jan 7 at 15:20












$begingroup$
ah... (x,y) are arguments here! Now that makes way more sense. Thanks. You should make an answer for that.
$endgroup$
– xotix
Jan 8 at 21:49




$begingroup$
ah... (x,y) are arguments here! Now that makes way more sense. Thanks. You should make an answer for that.
$endgroup$
– xotix
Jan 8 at 21:49










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