Change of variable in 2D integral
$begingroup$
If I have an integral of the form
$$
I = int_{-T}^T dxint_{-T}^T dy i(x - y)
tag1$$
Where $i$ is any function depending just on the relative variable, i.e., $x - y$. But, let's suppose that I need to perform the following change of variable:
$$
r = frac{x + y}{2}, quad R = x - y
tag2$$
Then, I think that $I$ will become
$$
I = int_{-sqrt{2}T/2}^{sqrt{2}T/2}dr int_{-sqrt{2}T}^{sqrt{2}T}dR i(R)
tag3$$
But I'm not sure. My reasoning to get this new limits is as follows: from Eq. (2), you can imagine $R$ as the module of a 2D vector in the cartesian plane, so
$$vec{R}_{maximum} = (T, 0) - (0, T) Rightarrow R_{maximum} = sqrt{2}T$$
Similarly,
$$R_{minimum} = -R_{maximum}$$
Moreover,
$$vec{r}_{maximum} = 0.5((T, 0) + (0, T)) Rightarrow r_{maximum} = frac{sqrt{2}T}{2}$$
and, as before,
$$r_{minimum} = -r_{maximum}$$
Is this correct or did I fail something?
calculus integration multivariable-calculus definite-integrals change-of-variable
asked Dec 23 '18 at 4:29
VickyVicky
2627
$endgroup$
add a comment |
$begingroup$
If I have an integral of the form
$$
I = int_{-T}^T dxint_{-T}^T dy i(x - y)
tag1$$
Where $i$ is any function depending just on the relative variable, i.e., $x - y$. But, let's suppose that I need to perform the following change of variable:
$$
r = frac{x + y}{2}, quad R = x - y
tag2$$
Then, I think that $I$ will become
$$
I = int_{-sqrt{2}T/2}^{sqrt{2}T/2}dr int_{-sqrt{2}T}^{sqrt{2}T}dR i(R)
tag3$$
But I'm not sure. My reasoning to get this new limits is as follows: from Eq. (2), you can imagine $R$ as the module of a 2D vector in the cartesian plane, so
$$vec{R}_{maximum} = (T, 0) - (0, T) Rightarrow R_{maximum} = sqrt{2}T$$
Similarly,
$$R_{minimum} = -R_{maximum}$$
Moreover,
$$vec{r}_{maximum} = 0.5((T, 0) + (0, T)) Rightarrow r_{maximum} = frac{sqrt{2}T}{2}$$
and, as before,
$$r_{minimum} = -r_{maximum}$$
Is this correct or did I fail something?
calculus integration multivariable-calculus definite-integrals change-of-variable
asked Dec 23 '18 at 4:29
VickyVicky
2627
$endgroup$
1
$begingroup$
Your change of variables maps a rectangle in the $(x,y)$-plane to a rhombus in the $(r,R)$-plane, not a rectangle. You need more elaborate limits on your new integral. I suggest drawing a picture.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 6:05
1
$begingroup$
The correct limits for an inner integral, after a change of variables, are not the largest and smallest possible values that variable can take ever; they are the largest and smallest possible values that variable can take given a particular value $r$ of the outer variable. In particular, the limits of the inner integral usually are functions of $r$ (or whatever the outer variable is).
$endgroup$
– Greg Martin
Dec 23 '18 at 6:18
add a comment |
$begingroup$
If I have an integral of the form
$$
I = int_{-T}^T dxint_{-T}^T dy i(x - y)
tag1$$
Where $i$ is any function depending just on the relative variable, i.e., $x - y$. But, let's suppose that I need to perform the following change of variable:
$$
r = frac{x + y}{2}, quad R = x - y
tag2$$
Then, I think that $I$ will become
$$
I = int_{-sqrt{2}T/2}^{sqrt{2}T/2}dr int_{-sqrt{2}T}^{sqrt{2}T}dR i(R)
tag3$$
But I'm not sure. My reasoning to get this new limits is as follows: from Eq. (2), you can imagine $R$ as the module of a 2D vector in the cartesian plane, so
$$vec{R}_{maximum} = (T, 0) - (0, T) Rightarrow R_{maximum} = sqrt{2}T$$
Similarly,
$$R_{minimum} = -R_{maximum}$$
Moreover,
$$vec{r}_{maximum} = 0.5((T, 0) + (0, T)) Rightarrow r_{maximum} = frac{sqrt{2}T}{2}$$
and, as before,
$$r_{minimum} = -r_{maximum}$$
Is this correct or did I fail something?
calculus integration multivariable-calculus definite-integrals change-of-variable
asked Dec 23 '18 at 4:29
VickyVicky
2627
$endgroup$
If I have an integral of the form
$$
I = int_{-T}^T dxint_{-T}^T dy i(x - y)
tag1$$
Where $i$ is any function depending just on the relative variable, i.e., $x - y$. But, let's suppose that I need to perform the following change of variable:
$$
r = frac{x + y}{2}, quad R = x - y
tag2$$
Then, I think that $I$ will become
$$
I = int_{-sqrt{2}T/2}^{sqrt{2}T/2}dr int_{-sqrt{2}T}^{sqrt{2}T}dR i(R)
tag3$$
But I'm not sure. My reasoning to get this new limits is as follows: from Eq. (2), you can imagine $R$ as the module of a 2D vector in the cartesian plane, so
$$vec{R}_{maximum} = (T, 0) - (0, T) Rightarrow R_{maximum} = sqrt{2}T$$
Similarly,
$$R_{minimum} = -R_{maximum}$$
Moreover,
$$vec{r}_{maximum} = 0.5((T, 0) + (0, T)) Rightarrow r_{maximum} = frac{sqrt{2}T}{2}$$
and, as before,
$$r_{minimum} = -r_{maximum}$$
Is this correct or did I fail something?
calculus integration multivariable-calculus definite-integrals change-of-variable
calculus integration multivariable-calculus definite-integrals change-of-variable
asked Dec 23 '18 at 4:29
VickyVicky
2627
asked Dec 23 '18 at 4:29
VickyVicky
2627
edited Dec 23 '18 at 4:54
Vicky
asked Dec 23 '18 at 4:29
VickyVicky
2627
asked Dec 23 '18 at 4:29
VickyVicky
2627
asked Dec 23 '18 at 4:29
VickyVicky
2627
2627
1
$begingroup$
Your change of variables maps a rectangle in the $(x,y)$-plane to a rhombus in the $(r,R)$-plane, not a rectangle. You need more elaborate limits on your new integral. I suggest drawing a picture.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 6:05
1
$begingroup$
The correct limits for an inner integral, after a change of variables, are not the largest and smallest possible values that variable can take ever; they are the largest and smallest possible values that variable can take given a particular value $r$ of the outer variable. In particular, the limits of the inner integral usually are functions of $r$ (or whatever the outer variable is).
$endgroup$
– Greg Martin
Dec 23 '18 at 6:18
add a comment |
1
$begingroup$
Your change of variables maps a rectangle in the $(x,y)$-plane to a rhombus in the $(r,R)$-plane, not a rectangle. You need more elaborate limits on your new integral. I suggest drawing a picture.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 6:05
1
$begingroup$
The correct limits for an inner integral, after a change of variables, are not the largest and smallest possible values that variable can take ever; they are the largest and smallest possible values that variable can take given a particular value $r$ of the outer variable. In particular, the limits of the inner integral usually are functions of $r$ (or whatever the outer variable is).
$endgroup$
– Greg Martin
Dec 23 '18 at 6:18
1
1
$begingroup$
Your change of variables maps a rectangle in the $(x,y)$-plane to a rhombus in the $(r,R)$-plane, not a rectangle. You need more elaborate limits on your new integral. I suggest drawing a picture.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 6:05
$begingroup$
Your change of variables maps a rectangle in the $(x,y)$-plane to a rhombus in the $(r,R)$-plane, not a rectangle. You need more elaborate limits on your new integral. I suggest drawing a picture.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 6:05
1
1
$begingroup$
The correct limits for an inner integral, after a change of variables, are not the largest and smallest possible values that variable can take ever; they are the largest and smallest possible values that variable can take given a particular value $r$ of the outer variable. In particular, the limits of the inner integral usually are functions of $r$ (or whatever the outer variable is).
$endgroup$
– Greg Martin
Dec 23 '18 at 6:18
$begingroup$
The correct limits for an inner integral, after a change of variables, are not the largest and smallest possible values that variable can take ever; they are the largest and smallest possible values that variable can take given a particular value $r$ of the outer variable. In particular, the limits of the inner integral usually are functions of $r$ (or whatever the outer variable is).
$endgroup$
– Greg Martin
Dec 23 '18 at 6:18
add a comment |
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$begingroup$
Your change of variables maps a rectangle in the $(x,y)$-plane to a rhombus in the $(r,R)$-plane, not a rectangle. You need more elaborate limits on your new integral. I suggest drawing a picture.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 6:05
1
$begingroup$
The correct limits for an inner integral, after a change of variables, are not the largest and smallest possible values that variable can take ever; they are the largest and smallest possible values that variable can take given a particular value $r$ of the outer variable. In particular, the limits of the inner integral usually are functions of $r$ (or whatever the outer variable is).
$endgroup$
– Greg Martin
Dec 23 '18 at 6:18