Mixing
Calculating a flux integral using Stokes vs. directly
$begingroup$
Let $G={(x,y,z)inmathbb R^3|x^2+y^2=1 , quad 0leq zleq 1}$
Let $f: mathbb R^3tomathbb R^3,quad f(x,y,z)=begin{pmatrix}yz^2\-x\ye^zend{pmatrix}$
Calculate $int_M curl(f)cdot n dS$ directly and with stokes. Consider the flow from inside to outside.
Solution:
We first see that we have a hollow cylnder with no bottom and no cap.
Stokes:
The boundary consists of the bottom and the cap's one. We parametrize both:
$gamma_B [2pi,0]tomathbb R^3, quad tmapsto begin{pmatrix}cos(t)\ sin(t) \ 0end{pmatrix} qquad dot{gamma_B}=begin{pmatrix}-sin(t)\ cos(t) \ 0end{pmatrix}$
$gamma_C [0,2pi]tomathbb R^3, quad tmapsto begin{pmatrix}cos(t)\ sin(t) \ 1end{pmatrix} qquad dot{gamma_C}=begin{pmatrix}-sin(t)\ cos(t) \ 0end{pmatrix}$
Now using stokes theorem we get
$int_M curl(f)cdot n dS=int_{gamma_B+gamma_C=partial M}f ds$
$=int_{2pi}^0begin{pmatrix}0\-cos(t)\ sin(t)end{pmatrix}cdotbegin{pmatrix}-sin(t)\ cos(t) \ 0end{pmatrix}dt+int_0^{2pi}begin{pmatrix}sin(t)\-cos(t)\ sin(t)end{pmatrix}cdotbegin{pmatrix}-sin(t)\ cos(t) \ 0end{pmatrix}dt$
$=int_{2pi}^0-cos^2(t)dt+int_0^{2pi}-sin^2(t)dt-cos^2(t)dt=pi-2pi=-pi$
Directly:
We parametrize the surface of $M$.
$Phi:[0,2pi]times [0,1]tomathbb R^3, quad (t,z)mapsto begin{pmatrix}cos(t)\ sin(t)\ zend{pmatrix}$
$Phi_ttimes Phi_z=begin{pmatrix}-sin(t)\ cos(t) \ 0end{pmatrix}times begin{pmatrix}0\0\ 1end{pmatrix}=begin{pmatrix}cos(t) \ sin(t) \ 0end{pmatrix}$
We see that $Phi_ttimes Phi_z$ is pointing in the correct direction. We calcualte the curl:
$curl(f)=begin{pmatrix}partial_x \ partial_y \ partial_zend{pmatrix}timesbegin{pmatrix}yz^2\-x\ye^zend{pmatrix} = begin{pmatrix}e^z \ 2yz \ -1-z^2end{pmatrix}$
$int_M curl(f)cdot n dS=int_0^{2pi}int_0^1 begin{pmatrix}e^z \ 2sin(t)z \ -1-z^2end{pmatrix} cdot begin{pmatrix}cos(t) \ sin(t) \ 0end{pmatrix} dzdt$
$=int_0^{2pi} int_0^1 e^zcos(t)+2sin^2(t)zdzdt$
$=underbrace{int_0^{2pi}cos(t)dt}_{=0}int_0^1 e^z dz + 2int_0^{2pi}sin^2(t)dtint_0^1zdz=0+frac{1}{2}2pi=pi$
Question: So as you can see, the two results don't match up and I'm not sure why.
integration stokes-theorem
edited Jan 26 at 10:57
Bernard
123k741117
asked Jan 26 at 10:12
xotixxotix
291411
$endgroup$
add a comment |
$begingroup$
Let $G={(x,y,z)inmathbb R^3|x^2+y^2=1 , quad 0leq zleq 1}$
Let $f: mathbb R^3tomathbb R^3,quad f(x,y,z)=begin{pmatrix}yz^2\-x\ye^zend{pmatrix}$
Calculate $int_M curl(f)cdot n dS$ directly and with stokes. Consider the flow from inside to outside.
Solution:
We first see that we have a hollow cylnder with no bottom and no cap.
Stokes:
The boundary consists of the bottom and the cap's one. We parametrize both:
$gamma_B [2pi,0]tomathbb R^3, quad tmapsto begin{pmatrix}cos(t)\ sin(t) \ 0end{pmatrix} qquad dot{gamma_B}=begin{pmatrix}-sin(t)\ cos(t) \ 0end{pmatrix}$
$gamma_C [0,2pi]tomathbb R^3, quad tmapsto begin{pmatrix}cos(t)\ sin(t) \ 1end{pmatrix} qquad dot{gamma_C}=begin{pmatrix}-sin(t)\ cos(t) \ 0end{pmatrix}$
Now using stokes theorem we get
$int_M curl(f)cdot n dS=int_{gamma_B+gamma_C=partial M}f ds$
$=int_{2pi}^0begin{pmatrix}0\-cos(t)\ sin(t)end{pmatrix}cdotbegin{pmatrix}-sin(t)\ cos(t) \ 0end{pmatrix}dt+int_0^{2pi}begin{pmatrix}sin(t)\-cos(t)\ sin(t)end{pmatrix}cdotbegin{pmatrix}-sin(t)\ cos(t) \ 0end{pmatrix}dt$
$=int_{2pi}^0-cos^2(t)dt+int_0^{2pi}-sin^2(t)dt-cos^2(t)dt=pi-2pi=-pi$
Directly:
We parametrize the surface of $M$.
$Phi:[0,2pi]times [0,1]tomathbb R^3, quad (t,z)mapsto begin{pmatrix}cos(t)\ sin(t)\ zend{pmatrix}$
$Phi_ttimes Phi_z=begin{pmatrix}-sin(t)\ cos(t) \ 0end{pmatrix}times begin{pmatrix}0\0\ 1end{pmatrix}=begin{pmatrix}cos(t) \ sin(t) \ 0end{pmatrix}$
We see that $Phi_ttimes Phi_z$ is pointing in the correct direction. We calcualte the curl:
$curl(f)=begin{pmatrix}partial_x \ partial_y \ partial_zend{pmatrix}timesbegin{pmatrix}yz^2\-x\ye^zend{pmatrix} = begin{pmatrix}e^z \ 2yz \ -1-z^2end{pmatrix}$
$int_M curl(f)cdot n dS=int_0^{2pi}int_0^1 begin{pmatrix}e^z \ 2sin(t)z \ -1-z^2end{pmatrix} cdot begin{pmatrix}cos(t) \ sin(t) \ 0end{pmatrix} dzdt$
$=int_0^{2pi} int_0^1 e^zcos(t)+2sin^2(t)zdzdt$
$=underbrace{int_0^{2pi}cos(t)dt}_{=0}int_0^1 e^z dz + 2int_0^{2pi}sin^2(t)dtint_0^1zdz=0+frac{1}{2}2pi=pi$
Question: So as you can see, the two results don't match up and I'm not sure why.
integration stokes-theorem
edited Jan 26 at 10:57
Bernard
123k741117
asked Jan 26 at 10:12
xotixxotix
291411
$endgroup$
add a comment |
$begingroup$
Let $G={(x,y,z)inmathbb R^3|x^2+y^2=1 , quad 0leq zleq 1}$
Let $f: mathbb R^3tomathbb R^3,quad f(x,y,z)=begin{pmatrix}yz^2\-x\ye^zend{pmatrix}$
Calculate $int_M curl(f)cdot n dS$ directly and with stokes. Consider the flow from inside to outside.
Solution:
We first see that we have a hollow cylnder with no bottom and no cap.
Stokes:
The boundary consists of the bottom and the cap's one. We parametrize both:
$gamma_B [2pi,0]tomathbb R^3, quad tmapsto begin{pmatrix}cos(t)\ sin(t) \ 0end{pmatrix} qquad dot{gamma_B}=begin{pmatrix}-sin(t)\ cos(t) \ 0end{pmatrix}$
$gamma_C [0,2pi]tomathbb R^3, quad tmapsto begin{pmatrix}cos(t)\ sin(t) \ 1end{pmatrix} qquad dot{gamma_C}=begin{pmatrix}-sin(t)\ cos(t) \ 0end{pmatrix}$
Now using stokes theorem we get
$int_M curl(f)cdot n dS=int_{gamma_B+gamma_C=partial M}f ds$
$=int_{2pi}^0begin{pmatrix}0\-cos(t)\ sin(t)end{pmatrix}cdotbegin{pmatrix}-sin(t)\ cos(t) \ 0end{pmatrix}dt+int_0^{2pi}begin{pmatrix}sin(t)\-cos(t)\ sin(t)end{pmatrix}cdotbegin{pmatrix}-sin(t)\ cos(t) \ 0end{pmatrix}dt$
$=int_{2pi}^0-cos^2(t)dt+int_0^{2pi}-sin^2(t)dt-cos^2(t)dt=pi-2pi=-pi$
Directly:
We parametrize the surface of $M$.
$Phi:[0,2pi]times [0,1]tomathbb R^3, quad (t,z)mapsto begin{pmatrix}cos(t)\ sin(t)\ zend{pmatrix}$
$Phi_ttimes Phi_z=begin{pmatrix}-sin(t)\ cos(t) \ 0end{pmatrix}times begin{pmatrix}0\0\ 1end{pmatrix}=begin{pmatrix}cos(t) \ sin(t) \ 0end{pmatrix}$
We see that $Phi_ttimes Phi_z$ is pointing in the correct direction. We calcualte the curl:
$curl(f)=begin{pmatrix}partial_x \ partial_y \ partial_zend{pmatrix}timesbegin{pmatrix}yz^2\-x\ye^zend{pmatrix} = begin{pmatrix}e^z \ 2yz \ -1-z^2end{pmatrix}$
$int_M curl(f)cdot n dS=int_0^{2pi}int_0^1 begin{pmatrix}e^z \ 2sin(t)z \ -1-z^2end{pmatrix} cdot begin{pmatrix}cos(t) \ sin(t) \ 0end{pmatrix} dzdt$
$=int_0^{2pi} int_0^1 e^zcos(t)+2sin^2(t)zdzdt$
$=underbrace{int_0^{2pi}cos(t)dt}_{=0}int_0^1 e^z dz + 2int_0^{2pi}sin^2(t)dtint_0^1zdz=0+frac{1}{2}2pi=pi$
Question: So as you can see, the two results don't match up and I'm not sure why.
integration stokes-theorem
edited Jan 26 at 10:57
Bernard
123k741117
asked Jan 26 at 10:12
xotixxotix
291411
$endgroup$
Let $G={(x,y,z)inmathbb R^3|x^2+y^2=1 , quad 0leq zleq 1}$
Let $f: mathbb R^3tomathbb R^3,quad f(x,y,z)=begin{pmatrix}yz^2\-x\ye^zend{pmatrix}$
Calculate $int_M curl(f)cdot n dS$ directly and with stokes. Consider the flow from inside to outside.
Solution:
We first see that we have a hollow cylnder with no bottom and no cap.
Stokes:
The boundary consists of the bottom and the cap's one. We parametrize both:
$gamma_B [2pi,0]tomathbb R^3, quad tmapsto begin{pmatrix}cos(t)\ sin(t) \ 0end{pmatrix} qquad dot{gamma_B}=begin{pmatrix}-sin(t)\ cos(t) \ 0end{pmatrix}$
$gamma_C [0,2pi]tomathbb R^3, quad tmapsto begin{pmatrix}cos(t)\ sin(t) \ 1end{pmatrix} qquad dot{gamma_C}=begin{pmatrix}-sin(t)\ cos(t) \ 0end{pmatrix}$
Now using stokes theorem we get
$int_M curl(f)cdot n dS=int_{gamma_B+gamma_C=partial M}f ds$
$=int_{2pi}^0begin{pmatrix}0\-cos(t)\ sin(t)end{pmatrix}cdotbegin{pmatrix}-sin(t)\ cos(t) \ 0end{pmatrix}dt+int_0^{2pi}begin{pmatrix}sin(t)\-cos(t)\ sin(t)end{pmatrix}cdotbegin{pmatrix}-sin(t)\ cos(t) \ 0end{pmatrix}dt$
$=int_{2pi}^0-cos^2(t)dt+int_0^{2pi}-sin^2(t)dt-cos^2(t)dt=pi-2pi=-pi$
Directly:
We parametrize the surface of $M$.
$Phi:[0,2pi]times [0,1]tomathbb R^3, quad (t,z)mapsto begin{pmatrix}cos(t)\ sin(t)\ zend{pmatrix}$
$Phi_ttimes Phi_z=begin{pmatrix}-sin(t)\ cos(t) \ 0end{pmatrix}times begin{pmatrix}0\0\ 1end{pmatrix}=begin{pmatrix}cos(t) \ sin(t) \ 0end{pmatrix}$
We see that $Phi_ttimes Phi_z$ is pointing in the correct direction. We calcualte the curl:
$curl(f)=begin{pmatrix}partial_x \ partial_y \ partial_zend{pmatrix}timesbegin{pmatrix}yz^2\-x\ye^zend{pmatrix} = begin{pmatrix}e^z \ 2yz \ -1-z^2end{pmatrix}$
$int_M curl(f)cdot n dS=int_0^{2pi}int_0^1 begin{pmatrix}e^z \ 2sin(t)z \ -1-z^2end{pmatrix} cdot begin{pmatrix}cos(t) \ sin(t) \ 0end{pmatrix} dzdt$
$=int_0^{2pi} int_0^1 e^zcos(t)+2sin^2(t)zdzdt$
$=underbrace{int_0^{2pi}cos(t)dt}_{=0}int_0^1 e^z dz + 2int_0^{2pi}sin^2(t)dtint_0^1zdz=0+frac{1}{2}2pi=pi$
Question: So as you can see, the two results don't match up and I'm not sure why.
integration stokes-theorem
integration stokes-theorem
edited Jan 26 at 10:57
Bernard
123k741117
asked Jan 26 at 10:12
xotixxotix
291411
edited Jan 26 at 10:57
Bernard
123k741117
asked Jan 26 at 10:12
xotixxotix
291411
edited Jan 26 at 10:57
Bernard
123k741117
edited Jan 26 at 10:57
Bernard
123k741117
edited Jan 26 at 10:57
Bernard
123k741117
123k741117
asked Jan 26 at 10:12
xotixxotix
291411
asked Jan 26 at 10:12
xotixxotix
291411
asked Jan 26 at 10:12
xotixxotix
291411
291411
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think that in order to campute the flow from inside to outside you should take the bottom circle $gamma_B$ counterclockwise and the cap circle $gamma_C$ clockwise:
$$=int^{2pi}_0-cos^2(t)dt+int^0_{2pi}-sin^2(t)dt-cos^2(t)dt=-pi+2pi=pi.$$
Thinking to a person that walks along a boundary curve in the direction given by its orientation, we should have that the head points
along the normal of the surface, and the left hand is over the surface.
answered Jan 26 at 10:27
Robert ZRobert Z
101k1070143
$endgroup$
$begingroup$
If I want the flow from inside to outside, the normal vector on the bottom looks downwards and the one from the cup looks upwards, no? So the persons stnds on top of the cylinder e.g. and walks coutnerclockwise around the border of the cup. No? Basically the right-thumb rule. Like: imgur.com/a/CNGNlQo
$endgroup$
– xotix
Jan 26 at 10:34
$begingroup$
You said that the surface is a "hollow cylnder with no bottom and no cap" so why are you considering the surfaces of those caps? See page 2 here: web.mit.edu/jorloff/www/18.01a-esg/notes/topic48.pdf
$endgroup$
– Robert Z
Jan 26 at 10:44
$begingroup$
I just relaized that's exaclty my problem! It's a hollow cylinder, so my surface is the part parallel to the z-axis! So yeah, you are right. Thanks
$endgroup$
– xotix
Jan 26 at 10:53
$begingroup$
@xotix Well done!!
$endgroup$
– Robert Z
Jan 26 at 10:59
$begingroup$
Maybe a follow up question: The parametrization of the contour is always mathematically positive when using stokes, right? It does not depend on the chosen direction of the normal field, right? So if I'd want the flux from outside to inside I could use the same parametriztion (since $vec{n}$ takes care of the considered direction of the flux, right?
$endgroup$
– xotix
Jan 26 at 11:02
|
show 1 more comment
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1 Answer
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1 Answer
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active
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votes
$begingroup$
I think that in order to campute the flow from inside to outside you should take the bottom circle $gamma_B$ counterclockwise and the cap circle $gamma_C$ clockwise:
$$=int^{2pi}_0-cos^2(t)dt+int^0_{2pi}-sin^2(t)dt-cos^2(t)dt=-pi+2pi=pi.$$
Thinking to a person that walks along a boundary curve in the direction given by its orientation, we should have that the head points
along the normal of the surface, and the left hand is over the surface.
answered Jan 26 at 10:27
Robert ZRobert Z
101k1070143
$endgroup$
$begingroup$
If I want the flow from inside to outside, the normal vector on the bottom looks downwards and the one from the cup looks upwards, no? So the persons stnds on top of the cylinder e.g. and walks coutnerclockwise around the border of the cup. No? Basically the right-thumb rule. Like: imgur.com/a/CNGNlQo
$endgroup$
– xotix
Jan 26 at 10:34
$begingroup$
You said that the surface is a "hollow cylnder with no bottom and no cap" so why are you considering the surfaces of those caps? See page 2 here: web.mit.edu/jorloff/www/18.01a-esg/notes/topic48.pdf
$endgroup$
– Robert Z
Jan 26 at 10:44
$begingroup$
I just relaized that's exaclty my problem! It's a hollow cylinder, so my surface is the part parallel to the z-axis! So yeah, you are right. Thanks
$endgroup$
– xotix
Jan 26 at 10:53
$begingroup$
@xotix Well done!!
$endgroup$
– Robert Z
Jan 26 at 10:59
$begingroup$
Maybe a follow up question: The parametrization of the contour is always mathematically positive when using stokes, right? It does not depend on the chosen direction of the normal field, right? So if I'd want the flux from outside to inside I could use the same parametriztion (since $vec{n}$ takes care of the considered direction of the flux, right?
$endgroup$
– xotix
Jan 26 at 11:02
|
show 1 more comment
$begingroup$
I think that in order to campute the flow from inside to outside you should take the bottom circle $gamma_B$ counterclockwise and the cap circle $gamma_C$ clockwise:
$$=int^{2pi}_0-cos^2(t)dt+int^0_{2pi}-sin^2(t)dt-cos^2(t)dt=-pi+2pi=pi.$$
Thinking to a person that walks along a boundary curve in the direction given by its orientation, we should have that the head points
along the normal of the surface, and the left hand is over the surface.
answered Jan 26 at 10:27
Robert ZRobert Z
101k1070143
$endgroup$
$begingroup$
If I want the flow from inside to outside, the normal vector on the bottom looks downwards and the one from the cup looks upwards, no? So the persons stnds on top of the cylinder e.g. and walks coutnerclockwise around the border of the cup. No? Basically the right-thumb rule. Like: imgur.com/a/CNGNlQo
$endgroup$
– xotix
Jan 26 at 10:34
$begingroup$
You said that the surface is a "hollow cylnder with no bottom and no cap" so why are you considering the surfaces of those caps? See page 2 here: web.mit.edu/jorloff/www/18.01a-esg/notes/topic48.pdf
$endgroup$
– Robert Z
Jan 26 at 10:44
$begingroup$
I just relaized that's exaclty my problem! It's a hollow cylinder, so my surface is the part parallel to the z-axis! So yeah, you are right. Thanks
$endgroup$
– xotix
Jan 26 at 10:53
$begingroup$
@xotix Well done!!
$endgroup$
– Robert Z
Jan 26 at 10:59
$begingroup$
Maybe a follow up question: The parametrization of the contour is always mathematically positive when using stokes, right? It does not depend on the chosen direction of the normal field, right? So if I'd want the flux from outside to inside I could use the same parametriztion (since $vec{n}$ takes care of the considered direction of the flux, right?
$endgroup$
– xotix
Jan 26 at 11:02
|
show 1 more comment
$begingroup$
I think that in order to campute the flow from inside to outside you should take the bottom circle $gamma_B$ counterclockwise and the cap circle $gamma_C$ clockwise:
$$=int^{2pi}_0-cos^2(t)dt+int^0_{2pi}-sin^2(t)dt-cos^2(t)dt=-pi+2pi=pi.$$
Thinking to a person that walks along a boundary curve in the direction given by its orientation, we should have that the head points
along the normal of the surface, and the left hand is over the surface.
answered Jan 26 at 10:27
Robert ZRobert Z
101k1070143
$endgroup$
I think that in order to campute the flow from inside to outside you should take the bottom circle $gamma_B$ counterclockwise and the cap circle $gamma_C$ clockwise:
$$=int^{2pi}_0-cos^2(t)dt+int^0_{2pi}-sin^2(t)dt-cos^2(t)dt=-pi+2pi=pi.$$
Thinking to a person that walks along a boundary curve in the direction given by its orientation, we should have that the head points
along the normal of the surface, and the left hand is over the surface.
answered Jan 26 at 10:27
Robert ZRobert Z
101k1070143
edited Jan 26 at 10:33
answered Jan 26 at 10:27
Robert ZRobert Z
101k1070143
answered Jan 26 at 10:27
Robert ZRobert Z
101k1070143
answered Jan 26 at 10:27
Robert ZRobert Z
101k1070143
101k1070143
$begingroup$
If I want the flow from inside to outside, the normal vector on the bottom looks downwards and the one from the cup looks upwards, no? So the persons stnds on top of the cylinder e.g. and walks coutnerclockwise around the border of the cup. No? Basically the right-thumb rule. Like: imgur.com/a/CNGNlQo
$endgroup$
– xotix
Jan 26 at 10:34
$begingroup$
You said that the surface is a "hollow cylnder with no bottom and no cap" so why are you considering the surfaces of those caps? See page 2 here: web.mit.edu/jorloff/www/18.01a-esg/notes/topic48.pdf
$endgroup$
– Robert Z
Jan 26 at 10:44
$begingroup$
I just relaized that's exaclty my problem! It's a hollow cylinder, so my surface is the part parallel to the z-axis! So yeah, you are right. Thanks
$endgroup$
– xotix
Jan 26 at 10:53
$begingroup$
@xotix Well done!!
$endgroup$
– Robert Z
Jan 26 at 10:59
$begingroup$
Maybe a follow up question: The parametrization of the contour is always mathematically positive when using stokes, right? It does not depend on the chosen direction of the normal field, right? So if I'd want the flux from outside to inside I could use the same parametriztion (since $vec{n}$ takes care of the considered direction of the flux, right?
$endgroup$
– xotix
Jan 26 at 11:02
|
show 1 more comment
$begingroup$
If I want the flow from inside to outside, the normal vector on the bottom looks downwards and the one from the cup looks upwards, no? So the persons stnds on top of the cylinder e.g. and walks coutnerclockwise around the border of the cup. No? Basically the right-thumb rule. Like: imgur.com/a/CNGNlQo
$endgroup$
– xotix
Jan 26 at 10:34
$begingroup$
You said that the surface is a "hollow cylnder with no bottom and no cap" so why are you considering the surfaces of those caps? See page 2 here: web.mit.edu/jorloff/www/18.01a-esg/notes/topic48.pdf
$endgroup$
– Robert Z
Jan 26 at 10:44
$begingroup$
I just relaized that's exaclty my problem! It's a hollow cylinder, so my surface is the part parallel to the z-axis! So yeah, you are right. Thanks
$endgroup$
– xotix
Jan 26 at 10:53
$begingroup$
@xotix Well done!!
$endgroup$
– Robert Z
Jan 26 at 10:59
$begingroup$
Maybe a follow up question: The parametrization of the contour is always mathematically positive when using stokes, right? It does not depend on the chosen direction of the normal field, right? So if I'd want the flux from outside to inside I could use the same parametriztion (since $vec{n}$ takes care of the considered direction of the flux, right?
$endgroup$
– xotix
Jan 26 at 11:02
$begingroup$
If I want the flow from inside to outside, the normal vector on the bottom looks downwards and the one from the cup looks upwards, no? So the persons stnds on top of the cylinder e.g. and walks coutnerclockwise around the border of the cup. No? Basically the right-thumb rule. Like: imgur.com/a/CNGNlQo
$endgroup$
– xotix
Jan 26 at 10:34
$begingroup$
If I want the flow from inside to outside, the normal vector on the bottom looks downwards and the one from the cup looks upwards, no? So the persons stnds on top of the cylinder e.g. and walks coutnerclockwise around the border of the cup. No? Basically the right-thumb rule. Like: imgur.com/a/CNGNlQo
$endgroup$
– xotix
Jan 26 at 10:34
$begingroup$
You said that the surface is a "hollow cylnder with no bottom and no cap" so why are you considering the surfaces of those caps? See page 2 here: web.mit.edu/jorloff/www/18.01a-esg/notes/topic48.pdf
$endgroup$
– Robert Z
Jan 26 at 10:44
$begingroup$
You said that the surface is a "hollow cylnder with no bottom and no cap" so why are you considering the surfaces of those caps? See page 2 here: web.mit.edu/jorloff/www/18.01a-esg/notes/topic48.pdf
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– Robert Z
Jan 26 at 10:44
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I just relaized that's exaclty my problem! It's a hollow cylinder, so my surface is the part parallel to the z-axis! So yeah, you are right. Thanks
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– xotix
Jan 26 at 10:53
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I just relaized that's exaclty my problem! It's a hollow cylinder, so my surface is the part parallel to the z-axis! So yeah, you are right. Thanks
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– xotix
Jan 26 at 10:53
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@xotix Well done!!
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– Robert Z
Jan 26 at 10:59
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@xotix Well done!!
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– Robert Z
Jan 26 at 10:59
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Maybe a follow up question: The parametrization of the contour is always mathematically positive when using stokes, right? It does not depend on the chosen direction of the normal field, right? So if I'd want the flux from outside to inside I could use the same parametriztion (since $vec{n}$ takes care of the considered direction of the flux, right?
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– xotix
Jan 26 at 11:02
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Maybe a follow up question: The parametrization of the contour is always mathematically positive when using stokes, right? It does not depend on the chosen direction of the normal field, right? So if I'd want the flux from outside to inside I could use the same parametriztion (since $vec{n}$ takes care of the considered direction of the flux, right?
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– xotix
Jan 26 at 11:02
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