determining content of a cross vault on a ceiling-verfication











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I found this question as an old exam exercise, which I would like help for solving it properly...I've been starring at it for quiet a long time now.



As the title says, it is about determining the content of a cross vault on a ceiling.



I think it is defined by two half cylindric surfaces with radius 1
on a square $ [-1,1] times [-1,1] $ which are ''going through'' each other. I hope you understand what I mean ^^



The equations for the cylindric surfaces are given by



$ x_1^2 + x_3^2=1 , 0 leq x_3 , |x_2| leq 1 $



$x_2^2 + x_3 ^2 = 1 , 0leq x_3, |x_1| leq 1 $



Any help very appreciated





My progress on this task is following:



fix $x_3 $
therefore you get
$x_1^2= 1-x_3^2 $ and $ x_2^2=1-x_3^2$



so $ - sqrt{1-x_3^2} leq x_1, x_2 leq sqrt{1-x_3^2}$



the surface on the square, will come to:



$ 4(1-x_3^2) $



And



$ int_0^1 4(1-x_3^2) = 8/3 $



what do you think?










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    up vote
    2
    down vote

    favorite
    2












    I found this question as an old exam exercise, which I would like help for solving it properly...I've been starring at it for quiet a long time now.



    As the title says, it is about determining the content of a cross vault on a ceiling.



    I think it is defined by two half cylindric surfaces with radius 1
    on a square $ [-1,1] times [-1,1] $ which are ''going through'' each other. I hope you understand what I mean ^^



    The equations for the cylindric surfaces are given by



    $ x_1^2 + x_3^2=1 , 0 leq x_3 , |x_2| leq 1 $



    $x_2^2 + x_3 ^2 = 1 , 0leq x_3, |x_1| leq 1 $



    Any help very appreciated





    My progress on this task is following:



    fix $x_3 $
    therefore you get
    $x_1^2= 1-x_3^2 $ and $ x_2^2=1-x_3^2$



    so $ - sqrt{1-x_3^2} leq x_1, x_2 leq sqrt{1-x_3^2}$



    the surface on the square, will come to:



    $ 4(1-x_3^2) $



    And



    $ int_0^1 4(1-x_3^2) = 8/3 $



    what do you think?










    share|cite|improve this question


























      up vote
      2
      down vote

      favorite
      2









      up vote
      2
      down vote

      favorite
      2






      2





      I found this question as an old exam exercise, which I would like help for solving it properly...I've been starring at it for quiet a long time now.



      As the title says, it is about determining the content of a cross vault on a ceiling.



      I think it is defined by two half cylindric surfaces with radius 1
      on a square $ [-1,1] times [-1,1] $ which are ''going through'' each other. I hope you understand what I mean ^^



      The equations for the cylindric surfaces are given by



      $ x_1^2 + x_3^2=1 , 0 leq x_3 , |x_2| leq 1 $



      $x_2^2 + x_3 ^2 = 1 , 0leq x_3, |x_1| leq 1 $



      Any help very appreciated





      My progress on this task is following:



      fix $x_3 $
      therefore you get
      $x_1^2= 1-x_3^2 $ and $ x_2^2=1-x_3^2$



      so $ - sqrt{1-x_3^2} leq x_1, x_2 leq sqrt{1-x_3^2}$



      the surface on the square, will come to:



      $ 4(1-x_3^2) $



      And



      $ int_0^1 4(1-x_3^2) = 8/3 $



      what do you think?










      share|cite|improve this question















      I found this question as an old exam exercise, which I would like help for solving it properly...I've been starring at it for quiet a long time now.



      As the title says, it is about determining the content of a cross vault on a ceiling.



      I think it is defined by two half cylindric surfaces with radius 1
      on a square $ [-1,1] times [-1,1] $ which are ''going through'' each other. I hope you understand what I mean ^^



      The equations for the cylindric surfaces are given by



      $ x_1^2 + x_3^2=1 , 0 leq x_3 , |x_2| leq 1 $



      $x_2^2 + x_3 ^2 = 1 , 0leq x_3, |x_1| leq 1 $



      Any help very appreciated





      My progress on this task is following:



      fix $x_3 $
      therefore you get
      $x_1^2= 1-x_3^2 $ and $ x_2^2=1-x_3^2$



      so $ - sqrt{1-x_3^2} leq x_1, x_2 leq sqrt{1-x_3^2}$



      the surface on the square, will come to:



      $ 4(1-x_3^2) $



      And



      $ int_0^1 4(1-x_3^2) = 8/3 $



      what do you think?







      real-analysis multivariable-calculus proof-verification multiple-integral






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      share|cite|improve this question













      share|cite|improve this question




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      edited Dec 5 at 19:31

























      asked Dec 3 at 20:58









      constant94

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