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?










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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?










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      $begingroup$


      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?










      share|cite|improve this question











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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






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      edited Jan 29 at 12:55









      YuiTo Cheng

      2,1862937




      2,1862937










      asked Jan 29 at 12:48









      spyerspyer

      1188




      1188






















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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?






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          • $begingroup$
            Yes! I do. Thank you!
            $endgroup$
            – spyer
            Jan 29 at 15:38












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          1 Answer
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          0












          $begingroup$

          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?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yes! I do. Thank you!
            $endgroup$
            – spyer
            Jan 29 at 15:38
















          0












          $begingroup$

          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?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yes! I do. Thank you!
            $endgroup$
            – spyer
            Jan 29 at 15:38














          0












          0








          0





          $begingroup$

          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?






          share|cite|improve this answer









          $endgroup$



          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?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 29 at 14:03









          rogerlrogerl

          18k22848




          18k22848












          • $begingroup$
            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




          $begingroup$
          Yes! I do. Thank you!
          $endgroup$
          – spyer
          Jan 29 at 15:38


















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