Verify Stokes' theorem on a hemisphere using a line integral
$begingroup$
When solving this question, I wrote $r(t)$ as $langlecos t,0,sin trangle$ then I differentiated it and got $dr=langle-sin t,0,cos trangle,dt$, then i substituted in the formula for line integrals. However, in the solution, $dr$ isn't like that?
Also, why did we assume $y = 0$?
surface-integrals line-integrals stokes-theorem
edited Dec 17 '18 at 14:37
user10354138
7,4322925
asked Dec 17 '18 at 14:04
Sundaze123Sundaze123
73
$endgroup$
|
show 1 more comment
$begingroup$
When solving this question, I wrote $r(t)$ as $langlecos t,0,sin trangle$ then I differentiated it and got $dr=langle-sin t,0,cos trangle,dt$, then i substituted in the formula for line integrals. However, in the solution, $dr$ isn't like that?
Also, why did we assume $y = 0$?
surface-integrals line-integrals stokes-theorem
edited Dec 17 '18 at 14:37
user10354138
7,4322925
asked Dec 17 '18 at 14:04
Sundaze123Sundaze123
73
$endgroup$
$begingroup$
I think you can go around the same curve in the other direction though, what do you get when you work it out?
$endgroup$
– Wesley Strik
Dec 17 '18 at 14:19
$begingroup$
I don't get why it'll differ in the parametrisation whether y is positive or negative.
$endgroup$
– Sundaze123
Dec 17 '18 at 14:19
$begingroup$
orientation in the direction of the positive y-axis r(t)=(sin(t),0,cos(t)) but in the negative I should use r(t)=(cos(t),0,sin(t))?
$endgroup$
– Sundaze123
Dec 17 '18 at 14:22
$begingroup$
You literally move around the curve in the opposite direction because you switch the role of $sin(t)$ and $cos(t)$
$endgroup$
– Wesley Strik
Dec 17 '18 at 18:48
$begingroup$
Yes, because you move around differently on your circular path $C$, we actually calculate the flux through the lower hemisphere, which is $pi$, which tells us that the flux enters from below into the sphere and then leaves from the upper half which gives a value of $-pi$. Naturally the amount entering should be the amount leaving the sphere, if we at least have conservation of flux.
$endgroup$
– Wesley Strik
Dec 17 '18 at 19:22
|
show 1 more comment
$begingroup$
When solving this question, I wrote $r(t)$ as $langlecos t,0,sin trangle$ then I differentiated it and got $dr=langle-sin t,0,cos trangle,dt$, then i substituted in the formula for line integrals. However, in the solution, $dr$ isn't like that?
Also, why did we assume $y = 0$?
surface-integrals line-integrals stokes-theorem
edited Dec 17 '18 at 14:37
user10354138
7,4322925
asked Dec 17 '18 at 14:04
Sundaze123Sundaze123
73
$endgroup$
When solving this question, I wrote $r(t)$ as $langlecos t,0,sin trangle$ then I differentiated it and got $dr=langle-sin t,0,cos trangle,dt$, then i substituted in the formula for line integrals. However, in the solution, $dr$ isn't like that?
Also, why did we assume $y = 0$?
surface-integrals line-integrals stokes-theorem
surface-integrals line-integrals stokes-theorem
edited Dec 17 '18 at 14:37
user10354138
7,4322925
asked Dec 17 '18 at 14:04
Sundaze123Sundaze123
73
edited Dec 17 '18 at 14:37
user10354138
7,4322925
asked Dec 17 '18 at 14:04
Sundaze123Sundaze123
73
edited Dec 17 '18 at 14:37
user10354138
7,4322925
edited Dec 17 '18 at 14:37
user10354138
7,4322925
edited Dec 17 '18 at 14:37
user10354138
7,4322925
7,4322925
asked Dec 17 '18 at 14:04
Sundaze123Sundaze123
73
asked Dec 17 '18 at 14:04
Sundaze123Sundaze123
73
asked Dec 17 '18 at 14:04
Sundaze123Sundaze123
73
73
$begingroup$
I think you can go around the same curve in the other direction though, what do you get when you work it out?
$endgroup$
– Wesley Strik
Dec 17 '18 at 14:19
$begingroup$
I don't get why it'll differ in the parametrisation whether y is positive or negative.
$endgroup$
– Sundaze123
Dec 17 '18 at 14:19
$begingroup$
orientation in the direction of the positive y-axis r(t)=(sin(t),0,cos(t)) but in the negative I should use r(t)=(cos(t),0,sin(t))?
$endgroup$
– Sundaze123
Dec 17 '18 at 14:22
$begingroup$
You literally move around the curve in the opposite direction because you switch the role of $sin(t)$ and $cos(t)$
$endgroup$
– Wesley Strik
Dec 17 '18 at 18:48
$begingroup$
Yes, because you move around differently on your circular path $C$, we actually calculate the flux through the lower hemisphere, which is $pi$, which tells us that the flux enters from below into the sphere and then leaves from the upper half which gives a value of $-pi$. Naturally the amount entering should be the amount leaving the sphere, if we at least have conservation of flux.
$endgroup$
– Wesley Strik
Dec 17 '18 at 19:22
|
show 1 more comment
$begingroup$
I think you can go around the same curve in the other direction though, what do you get when you work it out?
$endgroup$
– Wesley Strik
Dec 17 '18 at 14:19
$begingroup$
I don't get why it'll differ in the parametrisation whether y is positive or negative.
$endgroup$
– Sundaze123
Dec 17 '18 at 14:19
$begingroup$
orientation in the direction of the positive y-axis r(t)=(sin(t),0,cos(t)) but in the negative I should use r(t)=(cos(t),0,sin(t))?
$endgroup$
– Sundaze123
Dec 17 '18 at 14:22
$begingroup$
You literally move around the curve in the opposite direction because you switch the role of $sin(t)$ and $cos(t)$
$endgroup$
– Wesley Strik
Dec 17 '18 at 18:48
$begingroup$
Yes, because you move around differently on your circular path $C$, we actually calculate the flux through the lower hemisphere, which is $pi$, which tells us that the flux enters from below into the sphere and then leaves from the upper half which gives a value of $-pi$. Naturally the amount entering should be the amount leaving the sphere, if we at least have conservation of flux.
$endgroup$
– Wesley Strik
Dec 17 '18 at 19:22
$begingroup$
I think you can go around the same curve in the other direction though, what do you get when you work it out?
$endgroup$
– Wesley Strik
Dec 17 '18 at 14:19
$begingroup$
I think you can go around the same curve in the other direction though, what do you get when you work it out?
$endgroup$
– Wesley Strik
Dec 17 '18 at 14:19
$begingroup$
I don't get why it'll differ in the parametrisation whether y is positive or negative.
$endgroup$
– Sundaze123
Dec 17 '18 at 14:19
$begingroup$
I don't get why it'll differ in the parametrisation whether y is positive or negative.
$endgroup$
– Sundaze123
Dec 17 '18 at 14:19
$begingroup$
orientation in the direction of the positive y-axis r(t)=(sin(t),0,cos(t)) but in the negative I should use r(t)=(cos(t),0,sin(t))?
$endgroup$
– Sundaze123
Dec 17 '18 at 14:22
$begingroup$
orientation in the direction of the positive y-axis r(t)=(sin(t),0,cos(t)) but in the negative I should use r(t)=(cos(t),0,sin(t))?
$endgroup$
– Sundaze123
Dec 17 '18 at 14:22
$begingroup$
You literally move around the curve in the opposite direction because you switch the role of $sin(t)$ and $cos(t)$
$endgroup$
– Wesley Strik
Dec 17 '18 at 18:48
$begingroup$
You literally move around the curve in the opposite direction because you switch the role of $sin(t)$ and $cos(t)$
$endgroup$
– Wesley Strik
Dec 17 '18 at 18:48
$begingroup$
Yes, because you move around differently on your circular path $C$, we actually calculate the flux through the lower hemisphere, which is $pi$, which tells us that the flux enters from below into the sphere and then leaves from the upper half which gives a value of $-pi$. Naturally the amount entering should be the amount leaving the sphere, if we at least have conservation of flux.
$endgroup$
– Wesley Strik
Dec 17 '18 at 19:22
$begingroup$
Yes, because you move around differently on your circular path $C$, we actually calculate the flux through the lower hemisphere, which is $pi$, which tells us that the flux enters from below into the sphere and then leaves from the upper half which gives a value of $-pi$. Naturally the amount entering should be the amount leaving the sphere, if we at least have conservation of flux.
$endgroup$
– Wesley Strik
Dec 17 '18 at 19:22
|
show 1 more comment
1 Answer
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oldest
votes
$begingroup$
Simply put, Stokes'theorem tells us that if we are interested in calculating the flux through a certain orientable surface in Euclidean space, we can simply consider the flux on a contour around that surface to end up with the same value. The important part here is that we always keep the same orientation. We naturally assume anti-clockwise orientation belonging to the upward direction, this is often called a right-handed coordinate system.
The problem states that we need an orientation in the direction of the positive $y$-axis, this means that our curves will run in an anti-clockwise direction see:
http://tutorial.math.lamar.edu/Classes/CalcIII/StokesTheorem.aspx
This is often called the natural direction. When we consider what closed curves we can form on our surface $C$, we realise these will be the great circles on the hemisphere, it is easiest to just take the one that intersects the $xz$ plane:
This parametrisation is given by:
$$r(t)=(sin(t), 0, cos(t)) $$
where $0leq t leq 2pi$, so our begin point is $(0,0,1)$, we then move towards $(frac{1}{2}sqrt{2},0,frac{1}{2}sqrt{2}$) etc, before finally returning to $(0,0,1)$, your method runs in the other direction (and has a phase difference of $frac{pi}{2}$), let's see if this should make a difference. Call it:
$$s(t)=(cos(t), 0, sin(t)) $$
If we compute our line integrals with $r(t)$ and $r'(t)$:
$$
oint_{t=0} ^{2pi} (0, cos(t) , sin(t) ) cdot (cos(t),0,-sin(t))dt=$$ $$ oint_{t=0} ^{2pi} - sin^2(t)dt=-pi$$
And now via your method $s(t)$ and $s'(t)$:
$$
oint_{t=0} ^{2pi} (0, sin(t) , cos(t) ) cdot (-sin(t),0,cos(t))dt =oint_{t=0} ^{2pi} cos^2(t)dt = pi$$
So indeed the difference is just a minus sign because of reversed orientation. Your method actually calculates the flux ($pi$) that enters the lower hemisphere and apparently (logically) this should leave ($- pi$)the upper hemisphere.
answered Dec 17 '18 at 18:48
Wesley StrikWesley Strik
2,052423
$endgroup$
add a comment |
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$begingroup$
Simply put, Stokes'theorem tells us that if we are interested in calculating the flux through a certain orientable surface in Euclidean space, we can simply consider the flux on a contour around that surface to end up with the same value. The important part here is that we always keep the same orientation. We naturally assume anti-clockwise orientation belonging to the upward direction, this is often called a right-handed coordinate system.
The problem states that we need an orientation in the direction of the positive $y$-axis, this means that our curves will run in an anti-clockwise direction see:
http://tutorial.math.lamar.edu/Classes/CalcIII/StokesTheorem.aspx
This is often called the natural direction. When we consider what closed curves we can form on our surface $C$, we realise these will be the great circles on the hemisphere, it is easiest to just take the one that intersects the $xz$ plane:
This parametrisation is given by:
$$r(t)=(sin(t), 0, cos(t)) $$
where $0leq t leq 2pi$, so our begin point is $(0,0,1)$, we then move towards $(frac{1}{2}sqrt{2},0,frac{1}{2}sqrt{2}$) etc, before finally returning to $(0,0,1)$, your method runs in the other direction (and has a phase difference of $frac{pi}{2}$), let's see if this should make a difference. Call it:
$$s(t)=(cos(t), 0, sin(t)) $$
If we compute our line integrals with $r(t)$ and $r'(t)$:
$$
oint_{t=0} ^{2pi} (0, cos(t) , sin(t) ) cdot (cos(t),0,-sin(t))dt=$$ $$ oint_{t=0} ^{2pi} - sin^2(t)dt=-pi$$
And now via your method $s(t)$ and $s'(t)$:
$$
oint_{t=0} ^{2pi} (0, sin(t) , cos(t) ) cdot (-sin(t),0,cos(t))dt =oint_{t=0} ^{2pi} cos^2(t)dt = pi$$
So indeed the difference is just a minus sign because of reversed orientation. Your method actually calculates the flux ($pi$) that enters the lower hemisphere and apparently (logically) this should leave ($- pi$)the upper hemisphere.
answered Dec 17 '18 at 18:48
Wesley StrikWesley Strik
2,052423
$endgroup$
add a comment |
$begingroup$
Simply put, Stokes'theorem tells us that if we are interested in calculating the flux through a certain orientable surface in Euclidean space, we can simply consider the flux on a contour around that surface to end up with the same value. The important part here is that we always keep the same orientation. We naturally assume anti-clockwise orientation belonging to the upward direction, this is often called a right-handed coordinate system.
The problem states that we need an orientation in the direction of the positive $y$-axis, this means that our curves will run in an anti-clockwise direction see:
http://tutorial.math.lamar.edu/Classes/CalcIII/StokesTheorem.aspx
This is often called the natural direction. When we consider what closed curves we can form on our surface $C$, we realise these will be the great circles on the hemisphere, it is easiest to just take the one that intersects the $xz$ plane:
This parametrisation is given by:
$$r(t)=(sin(t), 0, cos(t)) $$
where $0leq t leq 2pi$, so our begin point is $(0,0,1)$, we then move towards $(frac{1}{2}sqrt{2},0,frac{1}{2}sqrt{2}$) etc, before finally returning to $(0,0,1)$, your method runs in the other direction (and has a phase difference of $frac{pi}{2}$), let's see if this should make a difference. Call it:
$$s(t)=(cos(t), 0, sin(t)) $$
If we compute our line integrals with $r(t)$ and $r'(t)$:
$$
oint_{t=0} ^{2pi} (0, cos(t) , sin(t) ) cdot (cos(t),0,-sin(t))dt=$$ $$ oint_{t=0} ^{2pi} - sin^2(t)dt=-pi$$
And now via your method $s(t)$ and $s'(t)$:
$$
oint_{t=0} ^{2pi} (0, sin(t) , cos(t) ) cdot (-sin(t),0,cos(t))dt =oint_{t=0} ^{2pi} cos^2(t)dt = pi$$
So indeed the difference is just a minus sign because of reversed orientation. Your method actually calculates the flux ($pi$) that enters the lower hemisphere and apparently (logically) this should leave ($- pi$)the upper hemisphere.
answered Dec 17 '18 at 18:48
Wesley StrikWesley Strik
2,052423
$endgroup$
add a comment |
$begingroup$
Simply put, Stokes'theorem tells us that if we are interested in calculating the flux through a certain orientable surface in Euclidean space, we can simply consider the flux on a contour around that surface to end up with the same value. The important part here is that we always keep the same orientation. We naturally assume anti-clockwise orientation belonging to the upward direction, this is often called a right-handed coordinate system.
The problem states that we need an orientation in the direction of the positive $y$-axis, this means that our curves will run in an anti-clockwise direction see:
http://tutorial.math.lamar.edu/Classes/CalcIII/StokesTheorem.aspx
This is often called the natural direction. When we consider what closed curves we can form on our surface $C$, we realise these will be the great circles on the hemisphere, it is easiest to just take the one that intersects the $xz$ plane:
This parametrisation is given by:
$$r(t)=(sin(t), 0, cos(t)) $$
where $0leq t leq 2pi$, so our begin point is $(0,0,1)$, we then move towards $(frac{1}{2}sqrt{2},0,frac{1}{2}sqrt{2}$) etc, before finally returning to $(0,0,1)$, your method runs in the other direction (and has a phase difference of $frac{pi}{2}$), let's see if this should make a difference. Call it:
$$s(t)=(cos(t), 0, sin(t)) $$
If we compute our line integrals with $r(t)$ and $r'(t)$:
$$
oint_{t=0} ^{2pi} (0, cos(t) , sin(t) ) cdot (cos(t),0,-sin(t))dt=$$ $$ oint_{t=0} ^{2pi} - sin^2(t)dt=-pi$$
And now via your method $s(t)$ and $s'(t)$:
$$
oint_{t=0} ^{2pi} (0, sin(t) , cos(t) ) cdot (-sin(t),0,cos(t))dt =oint_{t=0} ^{2pi} cos^2(t)dt = pi$$
So indeed the difference is just a minus sign because of reversed orientation. Your method actually calculates the flux ($pi$) that enters the lower hemisphere and apparently (logically) this should leave ($- pi$)the upper hemisphere.
answered Dec 17 '18 at 18:48
Wesley StrikWesley Strik
2,052423
$endgroup$
Simply put, Stokes'theorem tells us that if we are interested in calculating the flux through a certain orientable surface in Euclidean space, we can simply consider the flux on a contour around that surface to end up with the same value. The important part here is that we always keep the same orientation. We naturally assume anti-clockwise orientation belonging to the upward direction, this is often called a right-handed coordinate system.
The problem states that we need an orientation in the direction of the positive $y$-axis, this means that our curves will run in an anti-clockwise direction see:
http://tutorial.math.lamar.edu/Classes/CalcIII/StokesTheorem.aspx
This is often called the natural direction. When we consider what closed curves we can form on our surface $C$, we realise these will be the great circles on the hemisphere, it is easiest to just take the one that intersects the $xz$ plane:
This parametrisation is given by:
$$r(t)=(sin(t), 0, cos(t)) $$
where $0leq t leq 2pi$, so our begin point is $(0,0,1)$, we then move towards $(frac{1}{2}sqrt{2},0,frac{1}{2}sqrt{2}$) etc, before finally returning to $(0,0,1)$, your method runs in the other direction (and has a phase difference of $frac{pi}{2}$), let's see if this should make a difference. Call it:
$$s(t)=(cos(t), 0, sin(t)) $$
If we compute our line integrals with $r(t)$ and $r'(t)$:
$$
oint_{t=0} ^{2pi} (0, cos(t) , sin(t) ) cdot (cos(t),0,-sin(t))dt=$$ $$ oint_{t=0} ^{2pi} - sin^2(t)dt=-pi$$
And now via your method $s(t)$ and $s'(t)$:
$$
oint_{t=0} ^{2pi} (0, sin(t) , cos(t) ) cdot (-sin(t),0,cos(t))dt =oint_{t=0} ^{2pi} cos^2(t)dt = pi$$
So indeed the difference is just a minus sign because of reversed orientation. Your method actually calculates the flux ($pi$) that enters the lower hemisphere and apparently (logically) this should leave ($- pi$)the upper hemisphere.
answered Dec 17 '18 at 18:48
Wesley StrikWesley Strik
2,052423
edited Dec 17 '18 at 19:25
answered Dec 17 '18 at 18:48
Wesley StrikWesley Strik
2,052423
answered Dec 17 '18 at 18:48
Wesley StrikWesley Strik
2,052423
answered Dec 17 '18 at 18:48
Wesley StrikWesley Strik
2,052423
2,052423
add a comment |
add a comment |
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$begingroup$
I think you can go around the same curve in the other direction though, what do you get when you work it out?
$endgroup$
– Wesley Strik
Dec 17 '18 at 14:19
$begingroup$
I don't get why it'll differ in the parametrisation whether y is positive or negative.
$endgroup$
– Sundaze123
Dec 17 '18 at 14:19
$begingroup$
orientation in the direction of the positive y-axis r(t)=(sin(t),0,cos(t)) but in the negative I should use r(t)=(cos(t),0,sin(t))?
$endgroup$
– Sundaze123
Dec 17 '18 at 14:22
$begingroup$
You literally move around the curve in the opposite direction because you switch the role of $sin(t)$ and $cos(t)$
$endgroup$
– Wesley Strik
Dec 17 '18 at 18:48
$begingroup$
Yes, because you move around differently on your circular path $C$, we actually calculate the flux through the lower hemisphere, which is $pi$, which tells us that the flux enters from below into the sphere and then leaves from the upper half which gives a value of $-pi$. Naturally the amount entering should be the amount leaving the sphere, if we at least have conservation of flux.
$endgroup$
– Wesley Strik
Dec 17 '18 at 19:22