Mixing
Calculate region using jacobian determinant of substitution
$begingroup$
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?
calculus jacobian
asked Jan 7 at 14:55
xotixxotix
263311
$endgroup$
add a comment |
$begingroup$
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?
calculus jacobian
asked Jan 7 at 14:55
xotixxotix
263311
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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?
calculus jacobian
asked Jan 7 at 14:55
xotixxotix
263311
$endgroup$
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
calculus jacobian
asked Jan 7 at 14:55
xotixxotix
263311
asked Jan 7 at 14:55
xotixxotix
263311
asked Jan 7 at 14:55
xotixxotix
263311
asked Jan 7 at 14:55
xotixxotix
263311
asked Jan 7 at 14:55
xotixxotix
263311
263311
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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