Integral in cylindrical coordinates with only one differential and written basis?
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Let's say I have this integral in cylindrical coordinates:
$$int_0^pi 2zrhoê_rho dphi$$
How is this calculated? I need some help in understanding how to interpret this integral.
- Why is there a $ê_rho$ after $rho$, but not a $ê_z$ after $z$? Does the $ê_rho$ "hold" all of the $2zrho$ as the "coefficient"?
- Can it then be written as a vector then, ie. $int_0^pi (2zrho,0,0)dphi$?
- Why is there only a $dphi$? Since there's no $phi$ in the function (vector?), can the $2zp$ be treated as "constants" and put outside the integral? Would the answer to the integral then be $2pi zrho$ or $2pi zrho ê_rho$
?
Kind of blurry questions, but the answers to them might help me understand more about coordinate basis and integrals.
integration vector-analysis coordinate-systems cylindrical-coordinates
asked Aug 28 at 16:52
Defindun
296
add a comment |Â
up vote
1
down vote
favorite
Let's say I have this integral in cylindrical coordinates:
$$int_0^pi 2zrhoê_rho dphi$$
How is this calculated? I need some help in understanding how to interpret this integral.
- Why is there a $ê_rho$ after $rho$, but not a $ê_z$ after $z$? Does the $ê_rho$ "hold" all of the $2zrho$ as the "coefficient"?
- Can it then be written as a vector then, ie. $int_0^pi (2zrho,0,0)dphi$?
- Why is there only a $dphi$? Since there's no $phi$ in the function (vector?), can the $2zp$ be treated as "constants" and put outside the integral? Would the answer to the integral then be $2pi zrho$ or $2pi zrho ê_rho$
?
Kind of blurry questions, but the answers to them might help me understand more about coordinate basis and integrals.
integration vector-analysis coordinate-systems cylindrical-coordinates
asked Aug 28 at 16:52
Defindun
296
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let's say I have this integral in cylindrical coordinates:
$$int_0^pi 2zrhoê_rho dphi$$
How is this calculated? I need some help in understanding how to interpret this integral.
- Why is there a $ê_rho$ after $rho$, but not a $ê_z$ after $z$? Does the $ê_rho$ "hold" all of the $2zrho$ as the "coefficient"?
- Can it then be written as a vector then, ie. $int_0^pi (2zrho,0,0)dphi$?
- Why is there only a $dphi$? Since there's no $phi$ in the function (vector?), can the $2zp$ be treated as "constants" and put outside the integral? Would the answer to the integral then be $2pi zrho$ or $2pi zrho ê_rho$
?
Kind of blurry questions, but the answers to them might help me understand more about coordinate basis and integrals.
integration vector-analysis coordinate-systems cylindrical-coordinates
asked Aug 28 at 16:52
Defindun
296
Let's say I have this integral in cylindrical coordinates:
$$int_0^pi 2zrhoê_rho dphi$$
How is this calculated? I need some help in understanding how to interpret this integral.
- Why is there a $ê_rho$ after $rho$, but not a $ê_z$ after $z$? Does the $ê_rho$ "hold" all of the $2zrho$ as the "coefficient"?
- Can it then be written as a vector then, ie. $int_0^pi (2zrho,0,0)dphi$?
- Why is there only a $dphi$? Since there's no $phi$ in the function (vector?), can the $2zp$ be treated as "constants" and put outside the integral? Would the answer to the integral then be $2pi zrho$ or $2pi zrho ê_rho$
?
Kind of blurry questions, but the answers to them might help me understand more about coordinate basis and integrals.
integration vector-analysis coordinate-systems cylindrical-coordinates
asked Aug 28 at 16:52
Defindun
296
asked Aug 28 at 16:52
Defindun
296
asked Aug 28 at 16:52
Defindun
296
asked Aug 28 at 16:52
Defindun
296
296
add a comment |Â
add a comment |Â
1 Answer
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oldest
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up vote
1
down vote
accepted
The scalar factor $2zrho$ is independent of $phi$ and can be taken outside the integral, but the vector $hate_rho$ is actually a function of $phi$; at each point, it's the unit vector pointing in the $rho$ direction at that point, so in terms of its $xyz$ components it's given by
$$
hate_rho =
beginpmatrix
cosphi \
sinphi \
0
endpmatrix
.
$$
So what you have is the integral of a vector-valued function, and you can integrate it component-wise after writing $hate_rho$ as above.
answered Aug 28 at 17:09
Hans Lundmark
33.5k564109
Okay, I can see that connection. But why would I want to change ê_rho to cartesian coordinates, since the integral is for cylindrical coordinates and the differential is of phi? Is this because ê_rho is dependent on phi as the "position vector circulates around in the cylinder"? So the primitive function would then be 2zÃÂ(sin(phi), -cos(phi), 0)? I thought ê_rho was used to describe a vector instead of the usual (2zÃÂ, 0, 0). Could the integral be written like that as well? How would I then see that I need to convert the ê_rho to your above vector and then integrate component-wise.
– Defindun
Aug 28 at 18:25
@Defindun: When you integrate, you are “adding up infinitely many infinitesimal contributionsâ€Â. In this case, you have a bunch of vectors to add up, and if you want to get the grand total in the $x$-direction, you add up all the $x$-components of the contributing vectors. And the same for $y$ and $z$. This makes sense, since the $x$, $y$ and $z$ directions don't change – they are the same at all points. But you can't directly add up all the components in the $rho$ direction, because there is no global $rho$ direction – it changes from point to point.
– Hans Lundmark
Aug 28 at 19:46
Okay, thank you! So that is because this is a vector-valued function in cylindrical coordinates, which can be written as ∫ (2zÃÂ, 0, 0) dÕ as well? So, since its vector-valued, we have to use coordinates that remains the same, ie. are global? So we convert the "vector-direction" which depends on Õ (since that's the differential) and can put 2zàoutside as "constants"? My question then is for regular triple integrals for example, where we integrate over ÃÂ, Õ, and z as they are, so to speak? What are the unit vectors for these? It is after all cylindrical coordinates as well.
– Defindun
Aug 29 at 5:45
@Defindun: Notation such as $(a,b,c)$ is shorthand and has to be interpreted in context. If what you mean is $a hate_x + b hate_y + c hate_z$, then you can integrate component-wise, since what you are doing is $int (a hate_x + b hate_y + c hate_z = hate_x int a + hate_y int b + hate_z int c$, pulling the constant vectors outside of the integral. But if you mean $a hate_rho + b hate_phi + c hate_z$, then you can't integrate component-wise, since the vectors $hate_rho$ and $hate_phi$ aren't constant.
– Hans Lundmark
Aug 29 at 6:08
@Defindun: But regarding your question about triple integrals, I don't understand what you mean. It's just standard multivariable calculus, where you integrate a scalar-valued function. A change of variables converts the original region in $xyz$-space to a new region in $rthetaphi$-space, but it's still just an ordinary triple integral no matter what the coordinates are called.
– Hans Lundmark
Aug 29 at 6:10
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The scalar factor $2zrho$ is independent of $phi$ and can be taken outside the integral, but the vector $hate_rho$ is actually a function of $phi$; at each point, it's the unit vector pointing in the $rho$ direction at that point, so in terms of its $xyz$ components it's given by
$$
hate_rho =
beginpmatrix
cosphi \
sinphi \
0
endpmatrix
.
$$
So what you have is the integral of a vector-valued function, and you can integrate it component-wise after writing $hate_rho$ as above.
answered Aug 28 at 17:09
Hans Lundmark
33.5k564109
Okay, I can see that connection. But why would I want to change ê_rho to cartesian coordinates, since the integral is for cylindrical coordinates and the differential is of phi? Is this because ê_rho is dependent on phi as the "position vector circulates around in the cylinder"? So the primitive function would then be 2zÃÂ(sin(phi), -cos(phi), 0)? I thought ê_rho was used to describe a vector instead of the usual (2zÃÂ, 0, 0). Could the integral be written like that as well? How would I then see that I need to convert the ê_rho to your above vector and then integrate component-wise.
– Defindun
Aug 28 at 18:25
@Defindun: When you integrate, you are “adding up infinitely many infinitesimal contributionsâ€Â. In this case, you have a bunch of vectors to add up, and if you want to get the grand total in the $x$-direction, you add up all the $x$-components of the contributing vectors. And the same for $y$ and $z$. This makes sense, since the $x$, $y$ and $z$ directions don't change – they are the same at all points. But you can't directly add up all the components in the $rho$ direction, because there is no global $rho$ direction – it changes from point to point.
– Hans Lundmark
Aug 28 at 19:46
Okay, thank you! So that is because this is a vector-valued function in cylindrical coordinates, which can be written as ∫ (2zÃÂ, 0, 0) dÕ as well? So, since its vector-valued, we have to use coordinates that remains the same, ie. are global? So we convert the "vector-direction" which depends on Õ (since that's the differential) and can put 2zàoutside as "constants"? My question then is for regular triple integrals for example, where we integrate over ÃÂ, Õ, and z as they are, so to speak? What are the unit vectors for these? It is after all cylindrical coordinates as well.
– Defindun
Aug 29 at 5:45
@Defindun: Notation such as $(a,b,c)$ is shorthand and has to be interpreted in context. If what you mean is $a hate_x + b hate_y + c hate_z$, then you can integrate component-wise, since what you are doing is $int (a hate_x + b hate_y + c hate_z = hate_x int a + hate_y int b + hate_z int c$, pulling the constant vectors outside of the integral. But if you mean $a hate_rho + b hate_phi + c hate_z$, then you can't integrate component-wise, since the vectors $hate_rho$ and $hate_phi$ aren't constant.
– Hans Lundmark
Aug 29 at 6:08
@Defindun: But regarding your question about triple integrals, I don't understand what you mean. It's just standard multivariable calculus, where you integrate a scalar-valued function. A change of variables converts the original region in $xyz$-space to a new region in $rthetaphi$-space, but it's still just an ordinary triple integral no matter what the coordinates are called.
– Hans Lundmark
Aug 29 at 6:10
 |Â
show 4 more comments
up vote
1
down vote
accepted
The scalar factor $2zrho$ is independent of $phi$ and can be taken outside the integral, but the vector $hate_rho$ is actually a function of $phi$; at each point, it's the unit vector pointing in the $rho$ direction at that point, so in terms of its $xyz$ components it's given by
$$
hate_rho =
beginpmatrix
cosphi \
sinphi \
0
endpmatrix
.
$$
So what you have is the integral of a vector-valued function, and you can integrate it component-wise after writing $hate_rho$ as above.
answered Aug 28 at 17:09
Hans Lundmark
33.5k564109
Okay, I can see that connection. But why would I want to change ê_rho to cartesian coordinates, since the integral is for cylindrical coordinates and the differential is of phi? Is this because ê_rho is dependent on phi as the "position vector circulates around in the cylinder"? So the primitive function would then be 2zÃÂ(sin(phi), -cos(phi), 0)? I thought ê_rho was used to describe a vector instead of the usual (2zÃÂ, 0, 0). Could the integral be written like that as well? How would I then see that I need to convert the ê_rho to your above vector and then integrate component-wise.
– Defindun
Aug 28 at 18:25
@Defindun: When you integrate, you are “adding up infinitely many infinitesimal contributionsâ€Â. In this case, you have a bunch of vectors to add up, and if you want to get the grand total in the $x$-direction, you add up all the $x$-components of the contributing vectors. And the same for $y$ and $z$. This makes sense, since the $x$, $y$ and $z$ directions don't change – they are the same at all points. But you can't directly add up all the components in the $rho$ direction, because there is no global $rho$ direction – it changes from point to point.
– Hans Lundmark
Aug 28 at 19:46
Okay, thank you! So that is because this is a vector-valued function in cylindrical coordinates, which can be written as ∫ (2zÃÂ, 0, 0) dÕ as well? So, since its vector-valued, we have to use coordinates that remains the same, ie. are global? So we convert the "vector-direction" which depends on Õ (since that's the differential) and can put 2zàoutside as "constants"? My question then is for regular triple integrals for example, where we integrate over ÃÂ, Õ, and z as they are, so to speak? What are the unit vectors for these? It is after all cylindrical coordinates as well.
– Defindun
Aug 29 at 5:45
@Defindun: Notation such as $(a,b,c)$ is shorthand and has to be interpreted in context. If what you mean is $a hate_x + b hate_y + c hate_z$, then you can integrate component-wise, since what you are doing is $int (a hate_x + b hate_y + c hate_z = hate_x int a + hate_y int b + hate_z int c$, pulling the constant vectors outside of the integral. But if you mean $a hate_rho + b hate_phi + c hate_z$, then you can't integrate component-wise, since the vectors $hate_rho$ and $hate_phi$ aren't constant.
– Hans Lundmark
Aug 29 at 6:08
@Defindun: But regarding your question about triple integrals, I don't understand what you mean. It's just standard multivariable calculus, where you integrate a scalar-valued function. A change of variables converts the original region in $xyz$-space to a new region in $rthetaphi$-space, but it's still just an ordinary triple integral no matter what the coordinates are called.
– Hans Lundmark
Aug 29 at 6:10
 |Â
show 4 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The scalar factor $2zrho$ is independent of $phi$ and can be taken outside the integral, but the vector $hate_rho$ is actually a function of $phi$; at each point, it's the unit vector pointing in the $rho$ direction at that point, so in terms of its $xyz$ components it's given by
$$
hate_rho =
beginpmatrix
cosphi \
sinphi \
0
endpmatrix
.
$$
So what you have is the integral of a vector-valued function, and you can integrate it component-wise after writing $hate_rho$ as above.
answered Aug 28 at 17:09
Hans Lundmark
33.5k564109
The scalar factor $2zrho$ is independent of $phi$ and can be taken outside the integral, but the vector $hate_rho$ is actually a function of $phi$; at each point, it's the unit vector pointing in the $rho$ direction at that point, so in terms of its $xyz$ components it's given by
$$
hate_rho =
beginpmatrix
cosphi \
sinphi \
0
endpmatrix
.
$$
So what you have is the integral of a vector-valued function, and you can integrate it component-wise after writing $hate_rho$ as above.
answered Aug 28 at 17:09
Hans Lundmark
33.5k564109
answered Aug 28 at 17:09
Hans Lundmark
33.5k564109
answered Aug 28 at 17:09
Hans Lundmark
33.5k564109
answered Aug 28 at 17:09
Hans Lundmark
33.5k564109
33.5k564109
Okay, I can see that connection. But why would I want to change ê_rho to cartesian coordinates, since the integral is for cylindrical coordinates and the differential is of phi? Is this because ê_rho is dependent on phi as the "position vector circulates around in the cylinder"? So the primitive function would then be 2zÃÂ(sin(phi), -cos(phi), 0)? I thought ê_rho was used to describe a vector instead of the usual (2zÃÂ, 0, 0). Could the integral be written like that as well? How would I then see that I need to convert the ê_rho to your above vector and then integrate component-wise.
– Defindun
Aug 28 at 18:25
@Defindun: When you integrate, you are “adding up infinitely many infinitesimal contributionsâ€Â. In this case, you have a bunch of vectors to add up, and if you want to get the grand total in the $x$-direction, you add up all the $x$-components of the contributing vectors. And the same for $y$ and $z$. This makes sense, since the $x$, $y$ and $z$ directions don't change – they are the same at all points. But you can't directly add up all the components in the $rho$ direction, because there is no global $rho$ direction – it changes from point to point.
– Hans Lundmark
Aug 28 at 19:46
Okay, thank you! So that is because this is a vector-valued function in cylindrical coordinates, which can be written as ∫ (2zÃÂ, 0, 0) dÕ as well? So, since its vector-valued, we have to use coordinates that remains the same, ie. are global? So we convert the "vector-direction" which depends on Õ (since that's the differential) and can put 2zàoutside as "constants"? My question then is for regular triple integrals for example, where we integrate over ÃÂ, Õ, and z as they are, so to speak? What are the unit vectors for these? It is after all cylindrical coordinates as well.
– Defindun
Aug 29 at 5:45
@Defindun: Notation such as $(a,b,c)$ is shorthand and has to be interpreted in context. If what you mean is $a hate_x + b hate_y + c hate_z$, then you can integrate component-wise, since what you are doing is $int (a hate_x + b hate_y + c hate_z = hate_x int a + hate_y int b + hate_z int c$, pulling the constant vectors outside of the integral. But if you mean $a hate_rho + b hate_phi + c hate_z$, then you can't integrate component-wise, since the vectors $hate_rho$ and $hate_phi$ aren't constant.
– Hans Lundmark
Aug 29 at 6:08
@Defindun: But regarding your question about triple integrals, I don't understand what you mean. It's just standard multivariable calculus, where you integrate a scalar-valued function. A change of variables converts the original region in $xyz$-space to a new region in $rthetaphi$-space, but it's still just an ordinary triple integral no matter what the coordinates are called.
– Hans Lundmark
Aug 29 at 6:10
 |Â
show 4 more comments
Okay, I can see that connection. But why would I want to change ê_rho to cartesian coordinates, since the integral is for cylindrical coordinates and the differential is of phi? Is this because ê_rho is dependent on phi as the "position vector circulates around in the cylinder"? So the primitive function would then be 2zÃÂ(sin(phi), -cos(phi), 0)? I thought ê_rho was used to describe a vector instead of the usual (2zÃÂ, 0, 0). Could the integral be written like that as well? How would I then see that I need to convert the ê_rho to your above vector and then integrate component-wise.
– Defindun
Aug 28 at 18:25
@Defindun: When you integrate, you are “adding up infinitely many infinitesimal contributionsâ€Â. In this case, you have a bunch of vectors to add up, and if you want to get the grand total in the $x$-direction, you add up all the $x$-components of the contributing vectors. And the same for $y$ and $z$. This makes sense, since the $x$, $y$ and $z$ directions don't change – they are the same at all points. But you can't directly add up all the components in the $rho$ direction, because there is no global $rho$ direction – it changes from point to point.
– Hans Lundmark
Aug 28 at 19:46
Okay, thank you! So that is because this is a vector-valued function in cylindrical coordinates, which can be written as ∫ (2zÃÂ, 0, 0) dÕ as well? So, since its vector-valued, we have to use coordinates that remains the same, ie. are global? So we convert the "vector-direction" which depends on Õ (since that's the differential) and can put 2zàoutside as "constants"? My question then is for regular triple integrals for example, where we integrate over ÃÂ, Õ, and z as they are, so to speak? What are the unit vectors for these? It is after all cylindrical coordinates as well.
– Defindun
Aug 29 at 5:45
@Defindun: Notation such as $(a,b,c)$ is shorthand and has to be interpreted in context. If what you mean is $a hate_x + b hate_y + c hate_z$, then you can integrate component-wise, since what you are doing is $int (a hate_x + b hate_y + c hate_z = hate_x int a + hate_y int b + hate_z int c$, pulling the constant vectors outside of the integral. But if you mean $a hate_rho + b hate_phi + c hate_z$, then you can't integrate component-wise, since the vectors $hate_rho$ and $hate_phi$ aren't constant.
– Hans Lundmark
Aug 29 at 6:08
@Defindun: But regarding your question about triple integrals, I don't understand what you mean. It's just standard multivariable calculus, where you integrate a scalar-valued function. A change of variables converts the original region in $xyz$-space to a new region in $rthetaphi$-space, but it's still just an ordinary triple integral no matter what the coordinates are called.
– Hans Lundmark
Aug 29 at 6:10
Okay, I can see that connection. But why would I want to change ê_rho to cartesian coordinates, since the integral is for cylindrical coordinates and the differential is of phi? Is this because ê_rho is dependent on phi as the "position vector circulates around in the cylinder"? So the primitive function would then be 2zÃÂ(sin(phi), -cos(phi), 0)? I thought ê_rho was used to describe a vector instead of the usual (2zÃÂ, 0, 0). Could the integral be written like that as well? How would I then see that I need to convert the ê_rho to your above vector and then integrate component-wise.
– Defindun
Aug 28 at 18:25
Okay, I can see that connection. But why would I want to change ê_rho to cartesian coordinates, since the integral is for cylindrical coordinates and the differential is of phi? Is this because ê_rho is dependent on phi as the "position vector circulates around in the cylinder"? So the primitive function would then be 2zÃÂ(sin(phi), -cos(phi), 0)? I thought ê_rho was used to describe a vector instead of the usual (2zÃÂ, 0, 0). Could the integral be written like that as well? How would I then see that I need to convert the ê_rho to your above vector and then integrate component-wise.
– Defindun
Aug 28 at 18:25
@Defindun: When you integrate, you are “adding up infinitely many infinitesimal contributionsâ€Â. In this case, you have a bunch of vectors to add up, and if you want to get the grand total in the $x$-direction, you add up all the $x$-components of the contributing vectors. And the same for $y$ and $z$. This makes sense, since the $x$, $y$ and $z$ directions don't change – they are the same at all points. But you can't directly add up all the components in the $rho$ direction, because there is no global $rho$ direction – it changes from point to point.
– Hans Lundmark
Aug 28 at 19:46
@Defindun: When you integrate, you are “adding up infinitely many infinitesimal contributionsâ€Â. In this case, you have a bunch of vectors to add up, and if you want to get the grand total in the $x$-direction, you add up all the $x$-components of the contributing vectors. And the same for $y$ and $z$. This makes sense, since the $x$, $y$ and $z$ directions don't change – they are the same at all points. But you can't directly add up all the components in the $rho$ direction, because there is no global $rho$ direction – it changes from point to point.
– Hans Lundmark
Aug 28 at 19:46
Okay, thank you! So that is because this is a vector-valued function in cylindrical coordinates, which can be written as ∫ (2zÃÂ, 0, 0) dÕ as well? So, since its vector-valued, we have to use coordinates that remains the same, ie. are global? So we convert the "vector-direction" which depends on Õ (since that's the differential) and can put 2zàoutside as "constants"? My question then is for regular triple integrals for example, where we integrate over ÃÂ, Õ, and z as they are, so to speak? What are the unit vectors for these? It is after all cylindrical coordinates as well.
– Defindun
Aug 29 at 5:45
Okay, thank you! So that is because this is a vector-valued function in cylindrical coordinates, which can be written as ∫ (2zÃÂ, 0, 0) dÕ as well? So, since its vector-valued, we have to use coordinates that remains the same, ie. are global? So we convert the "vector-direction" which depends on Õ (since that's the differential) and can put 2zàoutside as "constants"? My question then is for regular triple integrals for example, where we integrate over ÃÂ, Õ, and z as they are, so to speak? What are the unit vectors for these? It is after all cylindrical coordinates as well.
– Defindun
Aug 29 at 5:45
@Defindun: Notation such as $(a,b,c)$ is shorthand and has to be interpreted in context. If what you mean is $a hate_x + b hate_y + c hate_z$, then you can integrate component-wise, since what you are doing is $int (a hate_x + b hate_y + c hate_z = hate_x int a + hate_y int b + hate_z int c$, pulling the constant vectors outside of the integral. But if you mean $a hate_rho + b hate_phi + c hate_z$, then you can't integrate component-wise, since the vectors $hate_rho$ and $hate_phi$ aren't constant.
– Hans Lundmark
Aug 29 at 6:08
@Defindun: Notation such as $(a,b,c)$ is shorthand and has to be interpreted in context. If what you mean is $a hate_x + b hate_y + c hate_z$, then you can integrate component-wise, since what you are doing is $int (a hate_x + b hate_y + c hate_z = hate_x int a + hate_y int b + hate_z int c$, pulling the constant vectors outside of the integral. But if you mean $a hate_rho + b hate_phi + c hate_z$, then you can't integrate component-wise, since the vectors $hate_rho$ and $hate_phi$ aren't constant.
– Hans Lundmark
Aug 29 at 6:08
@Defindun: But regarding your question about triple integrals, I don't understand what you mean. It's just standard multivariable calculus, where you integrate a scalar-valued function. A change of variables converts the original region in $xyz$-space to a new region in $rthetaphi$-space, but it's still just an ordinary triple integral no matter what the coordinates are called.
– Hans Lundmark
Aug 29 at 6:10
@Defindun: But regarding your question about triple integrals, I don't understand what you mean. It's just standard multivariable calculus, where you integrate a scalar-valued function. A change of variables converts the original region in $xyz$-space to a new region in $rthetaphi$-space, but it's still just an ordinary triple integral no matter what the coordinates are called.
– Hans Lundmark
Aug 29 at 6:10
 |Â
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