Are there any “sure” ways to tell what is a boundary for the purpose of Stokes' Theorem?
$begingroup$
If I had a region bounded by the surface $x^2 + y^2 = 9$ and a plane $x+y+z=1$, how would I determine what "boundary" I'd take the line integral over? Or would I take it over both boundaries?
The first being the circle of radius 3 with centre at the origin on the x-y plane
and the second, being the "ellipse-looking" boundary at the top (so the equation that comes out when we solve those two simultaneously)
?
Image of the region (so that small wedge above the x-y plane and below the surface plane): http://i.imgur.com/lim8gSZ.png?
Website used: https://www.geogebra.org/3d
stokes-theorem
$endgroup$
|
show 7 more comments
$begingroup$
If I had a region bounded by the surface $x^2 + y^2 = 9$ and a plane $x+y+z=1$, how would I determine what "boundary" I'd take the line integral over? Or would I take it over both boundaries?
The first being the circle of radius 3 with centre at the origin on the x-y plane
and the second, being the "ellipse-looking" boundary at the top (so the equation that comes out when we solve those two simultaneously)
?
Image of the region (so that small wedge above the x-y plane and below the surface plane): http://i.imgur.com/lim8gSZ.png?
Website used: https://www.geogebra.org/3d
stokes-theorem
$endgroup$
$begingroup$
Thanks. Why can't it also be the circle at the x-y plane? Iirc, I did a question that had to do with a hemisphere, laying on the x-y plane. We treated the boundary of this hemisphere surface to be that circle on the x-y plane. If we were to intersect this with a similar plane in the OP, would the new boundary be that intersection?
$endgroup$
– Twenty-six colours
May 30 '17 at 10:44
$begingroup$
The ellipse just isn't a subset of the $xy$-plane: As you can check, $(3, 0, -2)$ is on the circle. I'm not sure I understand the second question---perhaps you can specify exactly the surfaces?
$endgroup$
– Travis
May 30 '17 at 10:49
$begingroup$
My bad. I'll give explicit examples: So I did a question before that treated the surface $S$ where $S$ is the hemisphere: $x^2 + y^2 + z^2 = 1$ and $z geq 0$, and claimed that the boundary (for our line integral via Stoke's) is the circle $x^2 + y^2 = 1$ at $z=0$ (so the $x-y$ plane) and intuitively, this makes sense. However, what if we had say another surface that bounds the hemisphere and that intersected this hemisphere e.g. so the surface $x+y+z=1$? What will we say the boundary (to use for Stoke's) would be of this new surface (the one that is bounded by the hemisphere,..
$endgroup$
– Twenty-six colours
May 30 '17 at 11:06
$begingroup$
but is below that other surface)? Is it that intersection it makes with the hemisphere? Pic: i.imgur.com/m4Tof8P.png?1
$endgroup$
– Twenty-six colours
May 30 '17 at 11:07
$begingroup$
After you "explicit examples" I'm now not sure I even understood your original question the way you meant it to be interpreted. I don't know what you mean by "another surface that bounds the hemisphere".
$endgroup$
– Travis
May 30 '17 at 11:22
|
show 7 more comments
$begingroup$
If I had a region bounded by the surface $x^2 + y^2 = 9$ and a plane $x+y+z=1$, how would I determine what "boundary" I'd take the line integral over? Or would I take it over both boundaries?
The first being the circle of radius 3 with centre at the origin on the x-y plane
and the second, being the "ellipse-looking" boundary at the top (so the equation that comes out when we solve those two simultaneously)
?
Image of the region (so that small wedge above the x-y plane and below the surface plane): http://i.imgur.com/lim8gSZ.png?
Website used: https://www.geogebra.org/3d
stokes-theorem
$endgroup$
If I had a region bounded by the surface $x^2 + y^2 = 9$ and a plane $x+y+z=1$, how would I determine what "boundary" I'd take the line integral over? Or would I take it over both boundaries?
The first being the circle of radius 3 with centre at the origin on the x-y plane
and the second, being the "ellipse-looking" boundary at the top (so the equation that comes out when we solve those two simultaneously)
?
Image of the region (so that small wedge above the x-y plane and below the surface plane): http://i.imgur.com/lim8gSZ.png?
Website used: https://www.geogebra.org/3d
stokes-theorem
stokes-theorem
edited Dec 15 '18 at 4:58
anomaly
17.4k42664
17.4k42664
asked May 30 '17 at 9:54
Twenty-six coloursTwenty-six colours
196323
196323
$begingroup$
Thanks. Why can't it also be the circle at the x-y plane? Iirc, I did a question that had to do with a hemisphere, laying on the x-y plane. We treated the boundary of this hemisphere surface to be that circle on the x-y plane. If we were to intersect this with a similar plane in the OP, would the new boundary be that intersection?
$endgroup$
– Twenty-six colours
May 30 '17 at 10:44
$begingroup$
The ellipse just isn't a subset of the $xy$-plane: As you can check, $(3, 0, -2)$ is on the circle. I'm not sure I understand the second question---perhaps you can specify exactly the surfaces?
$endgroup$
– Travis
May 30 '17 at 10:49
$begingroup$
My bad. I'll give explicit examples: So I did a question before that treated the surface $S$ where $S$ is the hemisphere: $x^2 + y^2 + z^2 = 1$ and $z geq 0$, and claimed that the boundary (for our line integral via Stoke's) is the circle $x^2 + y^2 = 1$ at $z=0$ (so the $x-y$ plane) and intuitively, this makes sense. However, what if we had say another surface that bounds the hemisphere and that intersected this hemisphere e.g. so the surface $x+y+z=1$? What will we say the boundary (to use for Stoke's) would be of this new surface (the one that is bounded by the hemisphere,..
$endgroup$
– Twenty-six colours
May 30 '17 at 11:06
$begingroup$
but is below that other surface)? Is it that intersection it makes with the hemisphere? Pic: i.imgur.com/m4Tof8P.png?1
$endgroup$
– Twenty-six colours
May 30 '17 at 11:07
$begingroup$
After you "explicit examples" I'm now not sure I even understood your original question the way you meant it to be interpreted. I don't know what you mean by "another surface that bounds the hemisphere".
$endgroup$
– Travis
May 30 '17 at 11:22
|
show 7 more comments
$begingroup$
Thanks. Why can't it also be the circle at the x-y plane? Iirc, I did a question that had to do with a hemisphere, laying on the x-y plane. We treated the boundary of this hemisphere surface to be that circle on the x-y plane. If we were to intersect this with a similar plane in the OP, would the new boundary be that intersection?
$endgroup$
– Twenty-six colours
May 30 '17 at 10:44
$begingroup$
The ellipse just isn't a subset of the $xy$-plane: As you can check, $(3, 0, -2)$ is on the circle. I'm not sure I understand the second question---perhaps you can specify exactly the surfaces?
$endgroup$
– Travis
May 30 '17 at 10:49
$begingroup$
My bad. I'll give explicit examples: So I did a question before that treated the surface $S$ where $S$ is the hemisphere: $x^2 + y^2 + z^2 = 1$ and $z geq 0$, and claimed that the boundary (for our line integral via Stoke's) is the circle $x^2 + y^2 = 1$ at $z=0$ (so the $x-y$ plane) and intuitively, this makes sense. However, what if we had say another surface that bounds the hemisphere and that intersected this hemisphere e.g. so the surface $x+y+z=1$? What will we say the boundary (to use for Stoke's) would be of this new surface (the one that is bounded by the hemisphere,..
$endgroup$
– Twenty-six colours
May 30 '17 at 11:06
$begingroup$
but is below that other surface)? Is it that intersection it makes with the hemisphere? Pic: i.imgur.com/m4Tof8P.png?1
$endgroup$
– Twenty-six colours
May 30 '17 at 11:07
$begingroup$
After you "explicit examples" I'm now not sure I even understood your original question the way you meant it to be interpreted. I don't know what you mean by "another surface that bounds the hemisphere".
$endgroup$
– Travis
May 30 '17 at 11:22
$begingroup$
Thanks. Why can't it also be the circle at the x-y plane? Iirc, I did a question that had to do with a hemisphere, laying on the x-y plane. We treated the boundary of this hemisphere surface to be that circle on the x-y plane. If we were to intersect this with a similar plane in the OP, would the new boundary be that intersection?
$endgroup$
– Twenty-six colours
May 30 '17 at 10:44
$begingroup$
Thanks. Why can't it also be the circle at the x-y plane? Iirc, I did a question that had to do with a hemisphere, laying on the x-y plane. We treated the boundary of this hemisphere surface to be that circle on the x-y plane. If we were to intersect this with a similar plane in the OP, would the new boundary be that intersection?
$endgroup$
– Twenty-six colours
May 30 '17 at 10:44
$begingroup$
The ellipse just isn't a subset of the $xy$-plane: As you can check, $(3, 0, -2)$ is on the circle. I'm not sure I understand the second question---perhaps you can specify exactly the surfaces?
$endgroup$
– Travis
May 30 '17 at 10:49
$begingroup$
The ellipse just isn't a subset of the $xy$-plane: As you can check, $(3, 0, -2)$ is on the circle. I'm not sure I understand the second question---perhaps you can specify exactly the surfaces?
$endgroup$
– Travis
May 30 '17 at 10:49
$begingroup$
My bad. I'll give explicit examples: So I did a question before that treated the surface $S$ where $S$ is the hemisphere: $x^2 + y^2 + z^2 = 1$ and $z geq 0$, and claimed that the boundary (for our line integral via Stoke's) is the circle $x^2 + y^2 = 1$ at $z=0$ (so the $x-y$ plane) and intuitively, this makes sense. However, what if we had say another surface that bounds the hemisphere and that intersected this hemisphere e.g. so the surface $x+y+z=1$? What will we say the boundary (to use for Stoke's) would be of this new surface (the one that is bounded by the hemisphere,..
$endgroup$
– Twenty-six colours
May 30 '17 at 11:06
$begingroup$
My bad. I'll give explicit examples: So I did a question before that treated the surface $S$ where $S$ is the hemisphere: $x^2 + y^2 + z^2 = 1$ and $z geq 0$, and claimed that the boundary (for our line integral via Stoke's) is the circle $x^2 + y^2 = 1$ at $z=0$ (so the $x-y$ plane) and intuitively, this makes sense. However, what if we had say another surface that bounds the hemisphere and that intersected this hemisphere e.g. so the surface $x+y+z=1$? What will we say the boundary (to use for Stoke's) would be of this new surface (the one that is bounded by the hemisphere,..
$endgroup$
– Twenty-six colours
May 30 '17 at 11:06
$begingroup$
but is below that other surface)? Is it that intersection it makes with the hemisphere? Pic: i.imgur.com/m4Tof8P.png?1
$endgroup$
– Twenty-six colours
May 30 '17 at 11:07
$begingroup$
but is below that other surface)? Is it that intersection it makes with the hemisphere? Pic: i.imgur.com/m4Tof8P.png?1
$endgroup$
– Twenty-six colours
May 30 '17 at 11:07
$begingroup$
After you "explicit examples" I'm now not sure I even understood your original question the way you meant it to be interpreted. I don't know what you mean by "another surface that bounds the hemisphere".
$endgroup$
– Travis
May 30 '17 at 11:22
$begingroup$
After you "explicit examples" I'm now not sure I even understood your original question the way you meant it to be interpreted. I don't know what you mean by "another surface that bounds the hemisphere".
$endgroup$
– Travis
May 30 '17 at 11:22
|
show 7 more comments
2 Answers
2
active
oldest
votes
$begingroup$
I think there's a couple issues here. First, there is no region bounded by the cylinder $x^2+y^2=1$ and the slanted plane $x+y+z=1$. The two surfaces split $Bbb{R}^3$ into 4 unbounded regions:
- inside the cylinder and above the plane
- inside the cylinder and below the plane
- outside the cylinder and above the plane
- outside the cylinder and below the plane
Including the $xy$-plane $z=0$ creates a total of 10 regions, two of which are bounded. Adding the last condition that $zgeq 0$ finally clears up the region in question. Before we can even discuss what the boundary is, we need to have the region well-defined, so I will assume that the set we are referring to is
$$ Omega = {(x,y,z)inmathbb{R}^3 | (x^2+y^2leq 1)wedge(x+y+zleq 1)wedge(zgeq 0) } $$
From here, the components that make up the boundary are clear, and it brings up another issue. We are discussing a three-dimensional space, so we should have a two-dimensional boundary, which is consistent with what he have here. However, it is not consistent with use of the (Kelvin-)Stokes' Theorem, which is used on oriented two-dimensional surfaces with one-dimensional boundary. Without seeing the integral you intend to solve with this method, there's no way to answer the question, but this region that you seem to be defining isn't "compatible" with that form of Stokes' theorem (in the sense that the $int_Omega text{curl}Fcdot text dS = 0$ for any closed surface $Omega$).
Below, I will show how to find the boundary of the set $Omega$ independent of its use in integration.
So, the circle and ellipse curves will not be the entire boundary, since they are 1-dimensional curves. Nonetheless, they are part of the boundary. Just like the boundary of a solid cube is made up of faces (two-dimensional solid squares), edges (one-dimensional curves), and vertices (zero-dimensional points), we have the same situation for our solid here.
The faces are the truncated circular disk base, the truncated elliptical disk ceiling, and the cylindrical wall. We can get these two-dimensional boundary pieces by changing one inequalities in the definition of $Omega$ to an equality, and changing the rest to strict inequalities. So, we get the base by restricting our points to those in the $xy$-plane:
$$partialOmega_{text{base}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2< 1)wedge(x+y+z< 1)wedge(z= 0) }$$
Note that we let $z=0$, so we could have written $x+yleq 1$ explicitly for the second inequality. Similarly, we get the ceiling by restricting our points to those in the slanted plane:
$$partialOmega_{text{ceiling}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2< 1)wedge(x+y+z = 1)wedge(z> 0) }$$
and we get the wall by restricting our points to those on the cylinder:
$$partialOmega_{text{wall}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2= 1)wedge(x+y+z < 1)wedge(z> 0) }$$
Next, we can do a similar process to get the different 1-dimensional edge pieces by setting two inequalities to equalities at a time. This will give us three different edge pieces: the circular edge, the elliptical edge, and the linear edge. Explicitly,
$$partialOmega_{text{circular}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2= 1)wedge(x+y+z < 1)wedge(z= 0) } \
partialOmega_{text{elliptical}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2= 1)wedge(x+y+z = 1)wedge(z> 0) } \
partialOmega_{text{linear}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2< 1)wedge(x+y+z = 1)wedge(z= 0) }$$
Lastly, set all three inequalities to equality and you will get the two vertex points
$$partialOmega_{text{vertices}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2= 1)wedge(x+y+z =1)wedge(z=0) }$$
We can construct the full boundary by taking the union of these pieces:
$$partialOmega = bigcuplimits_{alpha} partialOmega_alpha $$
Note that this is a disjoint union; due to the use of strict inequalities mixed with equalities, it should be clear that no element of one piece could possibly be an element of another. This technique works in general for sets defined by inequalities like this, but you may find some pieces to be empty sets (such as the intersection of parallel faces of a cube).
$endgroup$
add a comment |
$begingroup$
You would take the line integral over each boundary separately then add the two results together. Be sure to orient the curves in the correct way so that the both correspond to the outward normal. In this instance the curve on the xy plane would be counterclockwise as viewed from above and the "ellipse-looking" boundary would be oriented clockwise. This is how we learned to approach it in my multivariable class but have only used it with cylinders.
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add a comment |
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2 Answers
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$begingroup$
I think there's a couple issues here. First, there is no region bounded by the cylinder $x^2+y^2=1$ and the slanted plane $x+y+z=1$. The two surfaces split $Bbb{R}^3$ into 4 unbounded regions:
- inside the cylinder and above the plane
- inside the cylinder and below the plane
- outside the cylinder and above the plane
- outside the cylinder and below the plane
Including the $xy$-plane $z=0$ creates a total of 10 regions, two of which are bounded. Adding the last condition that $zgeq 0$ finally clears up the region in question. Before we can even discuss what the boundary is, we need to have the region well-defined, so I will assume that the set we are referring to is
$$ Omega = {(x,y,z)inmathbb{R}^3 | (x^2+y^2leq 1)wedge(x+y+zleq 1)wedge(zgeq 0) } $$
From here, the components that make up the boundary are clear, and it brings up another issue. We are discussing a three-dimensional space, so we should have a two-dimensional boundary, which is consistent with what he have here. However, it is not consistent with use of the (Kelvin-)Stokes' Theorem, which is used on oriented two-dimensional surfaces with one-dimensional boundary. Without seeing the integral you intend to solve with this method, there's no way to answer the question, but this region that you seem to be defining isn't "compatible" with that form of Stokes' theorem (in the sense that the $int_Omega text{curl}Fcdot text dS = 0$ for any closed surface $Omega$).
Below, I will show how to find the boundary of the set $Omega$ independent of its use in integration.
So, the circle and ellipse curves will not be the entire boundary, since they are 1-dimensional curves. Nonetheless, they are part of the boundary. Just like the boundary of a solid cube is made up of faces (two-dimensional solid squares), edges (one-dimensional curves), and vertices (zero-dimensional points), we have the same situation for our solid here.
The faces are the truncated circular disk base, the truncated elliptical disk ceiling, and the cylindrical wall. We can get these two-dimensional boundary pieces by changing one inequalities in the definition of $Omega$ to an equality, and changing the rest to strict inequalities. So, we get the base by restricting our points to those in the $xy$-plane:
$$partialOmega_{text{base}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2< 1)wedge(x+y+z< 1)wedge(z= 0) }$$
Note that we let $z=0$, so we could have written $x+yleq 1$ explicitly for the second inequality. Similarly, we get the ceiling by restricting our points to those in the slanted plane:
$$partialOmega_{text{ceiling}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2< 1)wedge(x+y+z = 1)wedge(z> 0) }$$
and we get the wall by restricting our points to those on the cylinder:
$$partialOmega_{text{wall}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2= 1)wedge(x+y+z < 1)wedge(z> 0) }$$
Next, we can do a similar process to get the different 1-dimensional edge pieces by setting two inequalities to equalities at a time. This will give us three different edge pieces: the circular edge, the elliptical edge, and the linear edge. Explicitly,
$$partialOmega_{text{circular}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2= 1)wedge(x+y+z < 1)wedge(z= 0) } \
partialOmega_{text{elliptical}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2= 1)wedge(x+y+z = 1)wedge(z> 0) } \
partialOmega_{text{linear}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2< 1)wedge(x+y+z = 1)wedge(z= 0) }$$
Lastly, set all three inequalities to equality and you will get the two vertex points
$$partialOmega_{text{vertices}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2= 1)wedge(x+y+z =1)wedge(z=0) }$$
We can construct the full boundary by taking the union of these pieces:
$$partialOmega = bigcuplimits_{alpha} partialOmega_alpha $$
Note that this is a disjoint union; due to the use of strict inequalities mixed with equalities, it should be clear that no element of one piece could possibly be an element of another. This technique works in general for sets defined by inequalities like this, but you may find some pieces to be empty sets (such as the intersection of parallel faces of a cube).
$endgroup$
add a comment |
$begingroup$
I think there's a couple issues here. First, there is no region bounded by the cylinder $x^2+y^2=1$ and the slanted plane $x+y+z=1$. The two surfaces split $Bbb{R}^3$ into 4 unbounded regions:
- inside the cylinder and above the plane
- inside the cylinder and below the plane
- outside the cylinder and above the plane
- outside the cylinder and below the plane
Including the $xy$-plane $z=0$ creates a total of 10 regions, two of which are bounded. Adding the last condition that $zgeq 0$ finally clears up the region in question. Before we can even discuss what the boundary is, we need to have the region well-defined, so I will assume that the set we are referring to is
$$ Omega = {(x,y,z)inmathbb{R}^3 | (x^2+y^2leq 1)wedge(x+y+zleq 1)wedge(zgeq 0) } $$
From here, the components that make up the boundary are clear, and it brings up another issue. We are discussing a three-dimensional space, so we should have a two-dimensional boundary, which is consistent with what he have here. However, it is not consistent with use of the (Kelvin-)Stokes' Theorem, which is used on oriented two-dimensional surfaces with one-dimensional boundary. Without seeing the integral you intend to solve with this method, there's no way to answer the question, but this region that you seem to be defining isn't "compatible" with that form of Stokes' theorem (in the sense that the $int_Omega text{curl}Fcdot text dS = 0$ for any closed surface $Omega$).
Below, I will show how to find the boundary of the set $Omega$ independent of its use in integration.
So, the circle and ellipse curves will not be the entire boundary, since they are 1-dimensional curves. Nonetheless, they are part of the boundary. Just like the boundary of a solid cube is made up of faces (two-dimensional solid squares), edges (one-dimensional curves), and vertices (zero-dimensional points), we have the same situation for our solid here.
The faces are the truncated circular disk base, the truncated elliptical disk ceiling, and the cylindrical wall. We can get these two-dimensional boundary pieces by changing one inequalities in the definition of $Omega$ to an equality, and changing the rest to strict inequalities. So, we get the base by restricting our points to those in the $xy$-plane:
$$partialOmega_{text{base}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2< 1)wedge(x+y+z< 1)wedge(z= 0) }$$
Note that we let $z=0$, so we could have written $x+yleq 1$ explicitly for the second inequality. Similarly, we get the ceiling by restricting our points to those in the slanted plane:
$$partialOmega_{text{ceiling}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2< 1)wedge(x+y+z = 1)wedge(z> 0) }$$
and we get the wall by restricting our points to those on the cylinder:
$$partialOmega_{text{wall}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2= 1)wedge(x+y+z < 1)wedge(z> 0) }$$
Next, we can do a similar process to get the different 1-dimensional edge pieces by setting two inequalities to equalities at a time. This will give us three different edge pieces: the circular edge, the elliptical edge, and the linear edge. Explicitly,
$$partialOmega_{text{circular}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2= 1)wedge(x+y+z < 1)wedge(z= 0) } \
partialOmega_{text{elliptical}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2= 1)wedge(x+y+z = 1)wedge(z> 0) } \
partialOmega_{text{linear}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2< 1)wedge(x+y+z = 1)wedge(z= 0) }$$
Lastly, set all three inequalities to equality and you will get the two vertex points
$$partialOmega_{text{vertices}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2= 1)wedge(x+y+z =1)wedge(z=0) }$$
We can construct the full boundary by taking the union of these pieces:
$$partialOmega = bigcuplimits_{alpha} partialOmega_alpha $$
Note that this is a disjoint union; due to the use of strict inequalities mixed with equalities, it should be clear that no element of one piece could possibly be an element of another. This technique works in general for sets defined by inequalities like this, but you may find some pieces to be empty sets (such as the intersection of parallel faces of a cube).
$endgroup$
add a comment |
$begingroup$
I think there's a couple issues here. First, there is no region bounded by the cylinder $x^2+y^2=1$ and the slanted plane $x+y+z=1$. The two surfaces split $Bbb{R}^3$ into 4 unbounded regions:
- inside the cylinder and above the plane
- inside the cylinder and below the plane
- outside the cylinder and above the plane
- outside the cylinder and below the plane
Including the $xy$-plane $z=0$ creates a total of 10 regions, two of which are bounded. Adding the last condition that $zgeq 0$ finally clears up the region in question. Before we can even discuss what the boundary is, we need to have the region well-defined, so I will assume that the set we are referring to is
$$ Omega = {(x,y,z)inmathbb{R}^3 | (x^2+y^2leq 1)wedge(x+y+zleq 1)wedge(zgeq 0) } $$
From here, the components that make up the boundary are clear, and it brings up another issue. We are discussing a three-dimensional space, so we should have a two-dimensional boundary, which is consistent with what he have here. However, it is not consistent with use of the (Kelvin-)Stokes' Theorem, which is used on oriented two-dimensional surfaces with one-dimensional boundary. Without seeing the integral you intend to solve with this method, there's no way to answer the question, but this region that you seem to be defining isn't "compatible" with that form of Stokes' theorem (in the sense that the $int_Omega text{curl}Fcdot text dS = 0$ for any closed surface $Omega$).
Below, I will show how to find the boundary of the set $Omega$ independent of its use in integration.
So, the circle and ellipse curves will not be the entire boundary, since they are 1-dimensional curves. Nonetheless, they are part of the boundary. Just like the boundary of a solid cube is made up of faces (two-dimensional solid squares), edges (one-dimensional curves), and vertices (zero-dimensional points), we have the same situation for our solid here.
The faces are the truncated circular disk base, the truncated elliptical disk ceiling, and the cylindrical wall. We can get these two-dimensional boundary pieces by changing one inequalities in the definition of $Omega$ to an equality, and changing the rest to strict inequalities. So, we get the base by restricting our points to those in the $xy$-plane:
$$partialOmega_{text{base}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2< 1)wedge(x+y+z< 1)wedge(z= 0) }$$
Note that we let $z=0$, so we could have written $x+yleq 1$ explicitly for the second inequality. Similarly, we get the ceiling by restricting our points to those in the slanted plane:
$$partialOmega_{text{ceiling}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2< 1)wedge(x+y+z = 1)wedge(z> 0) }$$
and we get the wall by restricting our points to those on the cylinder:
$$partialOmega_{text{wall}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2= 1)wedge(x+y+z < 1)wedge(z> 0) }$$
Next, we can do a similar process to get the different 1-dimensional edge pieces by setting two inequalities to equalities at a time. This will give us three different edge pieces: the circular edge, the elliptical edge, and the linear edge. Explicitly,
$$partialOmega_{text{circular}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2= 1)wedge(x+y+z < 1)wedge(z= 0) } \
partialOmega_{text{elliptical}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2= 1)wedge(x+y+z = 1)wedge(z> 0) } \
partialOmega_{text{linear}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2< 1)wedge(x+y+z = 1)wedge(z= 0) }$$
Lastly, set all three inequalities to equality and you will get the two vertex points
$$partialOmega_{text{vertices}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2= 1)wedge(x+y+z =1)wedge(z=0) }$$
We can construct the full boundary by taking the union of these pieces:
$$partialOmega = bigcuplimits_{alpha} partialOmega_alpha $$
Note that this is a disjoint union; due to the use of strict inequalities mixed with equalities, it should be clear that no element of one piece could possibly be an element of another. This technique works in general for sets defined by inequalities like this, but you may find some pieces to be empty sets (such as the intersection of parallel faces of a cube).
$endgroup$
I think there's a couple issues here. First, there is no region bounded by the cylinder $x^2+y^2=1$ and the slanted plane $x+y+z=1$. The two surfaces split $Bbb{R}^3$ into 4 unbounded regions:
- inside the cylinder and above the plane
- inside the cylinder and below the plane
- outside the cylinder and above the plane
- outside the cylinder and below the plane
Including the $xy$-plane $z=0$ creates a total of 10 regions, two of which are bounded. Adding the last condition that $zgeq 0$ finally clears up the region in question. Before we can even discuss what the boundary is, we need to have the region well-defined, so I will assume that the set we are referring to is
$$ Omega = {(x,y,z)inmathbb{R}^3 | (x^2+y^2leq 1)wedge(x+y+zleq 1)wedge(zgeq 0) } $$
From here, the components that make up the boundary are clear, and it brings up another issue. We are discussing a three-dimensional space, so we should have a two-dimensional boundary, which is consistent with what he have here. However, it is not consistent with use of the (Kelvin-)Stokes' Theorem, which is used on oriented two-dimensional surfaces with one-dimensional boundary. Without seeing the integral you intend to solve with this method, there's no way to answer the question, but this region that you seem to be defining isn't "compatible" with that form of Stokes' theorem (in the sense that the $int_Omega text{curl}Fcdot text dS = 0$ for any closed surface $Omega$).
Below, I will show how to find the boundary of the set $Omega$ independent of its use in integration.
So, the circle and ellipse curves will not be the entire boundary, since they are 1-dimensional curves. Nonetheless, they are part of the boundary. Just like the boundary of a solid cube is made up of faces (two-dimensional solid squares), edges (one-dimensional curves), and vertices (zero-dimensional points), we have the same situation for our solid here.
The faces are the truncated circular disk base, the truncated elliptical disk ceiling, and the cylindrical wall. We can get these two-dimensional boundary pieces by changing one inequalities in the definition of $Omega$ to an equality, and changing the rest to strict inequalities. So, we get the base by restricting our points to those in the $xy$-plane:
$$partialOmega_{text{base}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2< 1)wedge(x+y+z< 1)wedge(z= 0) }$$
Note that we let $z=0$, so we could have written $x+yleq 1$ explicitly for the second inequality. Similarly, we get the ceiling by restricting our points to those in the slanted plane:
$$partialOmega_{text{ceiling}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2< 1)wedge(x+y+z = 1)wedge(z> 0) }$$
and we get the wall by restricting our points to those on the cylinder:
$$partialOmega_{text{wall}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2= 1)wedge(x+y+z < 1)wedge(z> 0) }$$
Next, we can do a similar process to get the different 1-dimensional edge pieces by setting two inequalities to equalities at a time. This will give us three different edge pieces: the circular edge, the elliptical edge, and the linear edge. Explicitly,
$$partialOmega_{text{circular}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2= 1)wedge(x+y+z < 1)wedge(z= 0) } \
partialOmega_{text{elliptical}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2= 1)wedge(x+y+z = 1)wedge(z> 0) } \
partialOmega_{text{linear}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2< 1)wedge(x+y+z = 1)wedge(z= 0) }$$
Lastly, set all three inequalities to equality and you will get the two vertex points
$$partialOmega_{text{vertices}} = {(x,y,z)inmathbb{R}^3 | (x^2+y^2= 1)wedge(x+y+z =1)wedge(z=0) }$$
We can construct the full boundary by taking the union of these pieces:
$$partialOmega = bigcuplimits_{alpha} partialOmega_alpha $$
Note that this is a disjoint union; due to the use of strict inequalities mixed with equalities, it should be clear that no element of one piece could possibly be an element of another. This technique works in general for sets defined by inequalities like this, but you may find some pieces to be empty sets (such as the intersection of parallel faces of a cube).
answered Dec 15 '18 at 6:12
AlexanderJ93AlexanderJ93
6,123823
6,123823
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You would take the line integral over each boundary separately then add the two results together. Be sure to orient the curves in the correct way so that the both correspond to the outward normal. In this instance the curve on the xy plane would be counterclockwise as viewed from above and the "ellipse-looking" boundary would be oriented clockwise. This is how we learned to approach it in my multivariable class but have only used it with cylinders.
$endgroup$
add a comment |
$begingroup$
You would take the line integral over each boundary separately then add the two results together. Be sure to orient the curves in the correct way so that the both correspond to the outward normal. In this instance the curve on the xy plane would be counterclockwise as viewed from above and the "ellipse-looking" boundary would be oriented clockwise. This is how we learned to approach it in my multivariable class but have only used it with cylinders.
$endgroup$
add a comment |
$begingroup$
You would take the line integral over each boundary separately then add the two results together. Be sure to orient the curves in the correct way so that the both correspond to the outward normal. In this instance the curve on the xy plane would be counterclockwise as viewed from above and the "ellipse-looking" boundary would be oriented clockwise. This is how we learned to approach it in my multivariable class but have only used it with cylinders.
$endgroup$
You would take the line integral over each boundary separately then add the two results together. Be sure to orient the curves in the correct way so that the both correspond to the outward normal. In this instance the curve on the xy plane would be counterclockwise as viewed from above and the "ellipse-looking" boundary would be oriented clockwise. This is how we learned to approach it in my multivariable class but have only used it with cylinders.
answered Dec 15 '18 at 4:56
Tim GreenTim Green
1
1
add a comment |
add a comment |
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$begingroup$
Thanks. Why can't it also be the circle at the x-y plane? Iirc, I did a question that had to do with a hemisphere, laying on the x-y plane. We treated the boundary of this hemisphere surface to be that circle on the x-y plane. If we were to intersect this with a similar plane in the OP, would the new boundary be that intersection?
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– Twenty-six colours
May 30 '17 at 10:44
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The ellipse just isn't a subset of the $xy$-plane: As you can check, $(3, 0, -2)$ is on the circle. I'm not sure I understand the second question---perhaps you can specify exactly the surfaces?
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– Travis
May 30 '17 at 10:49
$begingroup$
My bad. I'll give explicit examples: So I did a question before that treated the surface $S$ where $S$ is the hemisphere: $x^2 + y^2 + z^2 = 1$ and $z geq 0$, and claimed that the boundary (for our line integral via Stoke's) is the circle $x^2 + y^2 = 1$ at $z=0$ (so the $x-y$ plane) and intuitively, this makes sense. However, what if we had say another surface that bounds the hemisphere and that intersected this hemisphere e.g. so the surface $x+y+z=1$? What will we say the boundary (to use for Stoke's) would be of this new surface (the one that is bounded by the hemisphere,..
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– Twenty-six colours
May 30 '17 at 11:06
$begingroup$
but is below that other surface)? Is it that intersection it makes with the hemisphere? Pic: i.imgur.com/m4Tof8P.png?1
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– Twenty-six colours
May 30 '17 at 11:07
$begingroup$
After you "explicit examples" I'm now not sure I even understood your original question the way you meant it to be interpreted. I don't know what you mean by "another surface that bounds the hemisphere".
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– Travis
May 30 '17 at 11:22