Mixing
Manifold with boundary - finding the boundary
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I have the manifold with boundary $M:= lbrace (x_1,x_2,x_3) in mathbb R^3 : x_1geq 0, x_1^2+x_2^2+x_3^2=1rbrace cuplbrace (x_1,x_2,x_3) in mathbb R^3 : x_1= 0, x_1^2+x_2^2+x_3^2leq1rbrace$ and I need to find the boundary of this manifold. I think it is $lbrace (x_1,x_2,x_3) in mathbb R^n : x_1= 0, x_2^2+x_3^2=1rbrace$, the other option is that the boundary is the empty set? I think the first is right? Am I wrong?
calculus manifolds
edited Jan 29 at 12:55
YuiTo Cheng
2,1862937
asked Jan 29 at 12:48
spyerspyer
1188
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add a comment |
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I have the manifold with boundary $M:= lbrace (x_1,x_2,x_3) in mathbb R^3 : x_1geq 0, x_1^2+x_2^2+x_3^2=1rbrace cuplbrace (x_1,x_2,x_3) in mathbb R^3 : x_1= 0, x_1^2+x_2^2+x_3^2leq1rbrace$ and I need to find the boundary of this manifold. I think it is $lbrace (x_1,x_2,x_3) in mathbb R^n : x_1= 0, x_2^2+x_3^2=1rbrace$, the other option is that the boundary is the empty set? I think the first is right? Am I wrong?
calculus manifolds
edited Jan 29 at 12:55
YuiTo Cheng
2,1862937
asked Jan 29 at 12:48
spyerspyer
1188
$endgroup$
add a comment |
$begingroup$
I have the manifold with boundary $M:= lbrace (x_1,x_2,x_3) in mathbb R^3 : x_1geq 0, x_1^2+x_2^2+x_3^2=1rbrace cuplbrace (x_1,x_2,x_3) in mathbb R^3 : x_1= 0, x_1^2+x_2^2+x_3^2leq1rbrace$ and I need to find the boundary of this manifold. I think it is $lbrace (x_1,x_2,x_3) in mathbb R^n : x_1= 0, x_2^2+x_3^2=1rbrace$, the other option is that the boundary is the empty set? I think the first is right? Am I wrong?
calculus manifolds
edited Jan 29 at 12:55
YuiTo Cheng
2,1862937
asked Jan 29 at 12:48
spyerspyer
1188
$endgroup$
I have the manifold with boundary $M:= lbrace (x_1,x_2,x_3) in mathbb R^3 : x_1geq 0, x_1^2+x_2^2+x_3^2=1rbrace cuplbrace (x_1,x_2,x_3) in mathbb R^3 : x_1= 0, x_1^2+x_2^2+x_3^2leq1rbrace$ and I need to find the boundary of this manifold. I think it is $lbrace (x_1,x_2,x_3) in mathbb R^n : x_1= 0, x_2^2+x_3^2=1rbrace$, the other option is that the boundary is the empty set? I think the first is right? Am I wrong?
calculus manifolds
calculus manifolds
edited Jan 29 at 12:55
YuiTo Cheng
2,1862937
asked Jan 29 at 12:48
spyerspyer
1188
edited Jan 29 at 12:55
YuiTo Cheng
2,1862937
asked Jan 29 at 12:48
spyerspyer
1188
edited Jan 29 at 12:55
YuiTo Cheng
2,1862937
edited Jan 29 at 12:55
YuiTo Cheng
2,1862937
edited Jan 29 at 12:55
YuiTo Cheng
2,1862937
2,1862937
asked Jan 29 at 12:48
spyerspyer
1188
asked Jan 29 at 12:48
spyerspyer
1188
asked Jan 29 at 12:48
spyerspyer
1188
1188
add a comment |
add a comment |
1 Answer
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The first set is a hemisphere sitting on the $x_2x_3$-plane, including the boundary of the hemisphere at $x_1=0$. The second set is a disk of radius $1$ in the $x_2x_3$-plane. Can you see that the resulting surface has no boundary?
answered Jan 29 at 14:03
rogerlrogerl
18k22848
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$begingroup$
Yes! I do. Thank you!
$endgroup$
– spyer
Jan 29 at 15:38
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The first set is a hemisphere sitting on the $x_2x_3$-plane, including the boundary of the hemisphere at $x_1=0$. The second set is a disk of radius $1$ in the $x_2x_3$-plane. Can you see that the resulting surface has no boundary?
answered Jan 29 at 14:03
rogerlrogerl
18k22848
$endgroup$
$begingroup$
Yes! I do. Thank you!
$endgroup$
– spyer
Jan 29 at 15:38
add a comment |
$begingroup$
The first set is a hemisphere sitting on the $x_2x_3$-plane, including the boundary of the hemisphere at $x_1=0$. The second set is a disk of radius $1$ in the $x_2x_3$-plane. Can you see that the resulting surface has no boundary?
answered Jan 29 at 14:03
rogerlrogerl
18k22848
$endgroup$
$begingroup$
Yes! I do. Thank you!
$endgroup$
– spyer
Jan 29 at 15:38
add a comment |
$begingroup$
The first set is a hemisphere sitting on the $x_2x_3$-plane, including the boundary of the hemisphere at $x_1=0$. The second set is a disk of radius $1$ in the $x_2x_3$-plane. Can you see that the resulting surface has no boundary?
answered Jan 29 at 14:03
rogerlrogerl
18k22848
$endgroup$
The first set is a hemisphere sitting on the $x_2x_3$-plane, including the boundary of the hemisphere at $x_1=0$. The second set is a disk of radius $1$ in the $x_2x_3$-plane. Can you see that the resulting surface has no boundary?
answered Jan 29 at 14:03
rogerlrogerl
18k22848
answered Jan 29 at 14:03
rogerlrogerl
18k22848
answered Jan 29 at 14:03
rogerlrogerl
18k22848
answered Jan 29 at 14:03
rogerlrogerl
18k22848
18k22848
$begingroup$
Yes! I do. Thank you!
$endgroup$
– spyer
Jan 29 at 15:38
add a comment |
$begingroup$
Yes! I do. Thank you!
$endgroup$
– spyer
Jan 29 at 15:38
$begingroup$
Yes! I do. Thank you!
$endgroup$
– spyer
Jan 29 at 15:38
$begingroup$
Yes! I do. Thank you!
$endgroup$
– spyer
Jan 29 at 15:38
add a comment |
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