Error in calculation of intersection of three cylinders












0












$begingroup$


Set $A:={(x,y,z)in mathbb R^{3}: x^2+y^2leq 1, x^2+z^2leq 1, y^2+z^2 leq 1}$



I want to find the volume. I have to use symmetry and without polar coordinates.



My idea: Using symmetry we can look at the first octant and can restrict $x,y,z geq 0$



$lambda^{d}(A)=int_{A}dxdydz$



And we then set $0leq zleq 1$ and $0 leq x leq sqrt{1-z^2}$ and $0 leq y leq sqrt{1-z^2}$



So $int_{A}dxdydz=int_{0}^{1}int_{0}^{sqrt{1-z^2}}int_{0}^{sqrt{1-z^2}}dxdydz=int_{0}^{1}2sqrt{1-z^2}dz$



And then I would use substitution, namely, $z = sin{x}Rightarrow dz =cos{x}dx$, so $int_{0}^{1}2sqrt{1-z^2}dz=int_{0}^{frac{pi}{2}}2sqrt{1-sin{x}^2}cos{x}dx=2int_{0}^{frac{pi}{2}}cos^{2}{x}dx=2int_{0}^{frac{pi}{2}}frac{1}{2}+frac{1}{2}cos{2x}dx=int_{0}^{frac{pi}{2}}1+cos{2x}dx=x+frac{sin{2x}}{2}vert_{0}^{frac{pi}{2}}=frac{pi}{2}$



So getting back to eight octants, we'd get $lambda^{d}(A)=4pi$ but I have been told this is incorrect. I do not understand where I went wrong






real-analysis integration measure-theory







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    0












    $begingroup$


    Set $A:={(x,y,z)in mathbb R^{3}: x^2+y^2leq 1, x^2+z^2leq 1, y^2+z^2 leq 1}$



    I want to find the volume. I have to use symmetry and without polar coordinates.



    My idea: Using symmetry we can look at the first octant and can restrict $x,y,z geq 0$



    $lambda^{d}(A)=int_{A}dxdydz$



    And we then set $0leq zleq 1$ and $0 leq x leq sqrt{1-z^2}$ and $0 leq y leq sqrt{1-z^2}$



    So $int_{A}dxdydz=int_{0}^{1}int_{0}^{sqrt{1-z^2}}int_{0}^{sqrt{1-z^2}}dxdydz=int_{0}^{1}2sqrt{1-z^2}dz$



    And then I would use substitution, namely, $z = sin{x}Rightarrow dz =cos{x}dx$, so $int_{0}^{1}2sqrt{1-z^2}dz=int_{0}^{frac{pi}{2}}2sqrt{1-sin{x}^2}cos{x}dx=2int_{0}^{frac{pi}{2}}cos^{2}{x}dx=2int_{0}^{frac{pi}{2}}frac{1}{2}+frac{1}{2}cos{2x}dx=int_{0}^{frac{pi}{2}}1+cos{2x}dx=x+frac{sin{2x}}{2}vert_{0}^{frac{pi}{2}}=frac{pi}{2}$



    So getting back to eight octants, we'd get $lambda^{d}(A)=4pi$ but I have been told this is incorrect. I do not understand where I went wrong






    real-analysis integration measure-theory







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Set $A:={(x,y,z)in mathbb R^{3}: x^2+y^2leq 1, x^2+z^2leq 1, y^2+z^2 leq 1}$



      I want to find the volume. I have to use symmetry and without polar coordinates.



      My idea: Using symmetry we can look at the first octant and can restrict $x,y,z geq 0$



      $lambda^{d}(A)=int_{A}dxdydz$



      And we then set $0leq zleq 1$ and $0 leq x leq sqrt{1-z^2}$ and $0 leq y leq sqrt{1-z^2}$



      So $int_{A}dxdydz=int_{0}^{1}int_{0}^{sqrt{1-z^2}}int_{0}^{sqrt{1-z^2}}dxdydz=int_{0}^{1}2sqrt{1-z^2}dz$



      And then I would use substitution, namely, $z = sin{x}Rightarrow dz =cos{x}dx$, so $int_{0}^{1}2sqrt{1-z^2}dz=int_{0}^{frac{pi}{2}}2sqrt{1-sin{x}^2}cos{x}dx=2int_{0}^{frac{pi}{2}}cos^{2}{x}dx=2int_{0}^{frac{pi}{2}}frac{1}{2}+frac{1}{2}cos{2x}dx=int_{0}^{frac{pi}{2}}1+cos{2x}dx=x+frac{sin{2x}}{2}vert_{0}^{frac{pi}{2}}=frac{pi}{2}$



      So getting back to eight octants, we'd get $lambda^{d}(A)=4pi$ but I have been told this is incorrect. I do not understand where I went wrong






      real-analysis integration measure-theory







      share|cite|improve this question









      $endgroup$




      Set $A:={(x,y,z)in mathbb R^{3}: x^2+y^2leq 1, x^2+z^2leq 1, y^2+z^2 leq 1}$



      I want to find the volume. I have to use symmetry and without polar coordinates.



      My idea: Using symmetry we can look at the first octant and can restrict $x,y,z geq 0$



      $lambda^{d}(A)=int_{A}dxdydz$



      And we then set $0leq zleq 1$ and $0 leq x leq sqrt{1-z^2}$ and $0 leq y leq sqrt{1-z^2}$



      So $int_{A}dxdydz=int_{0}^{1}int_{0}^{sqrt{1-z^2}}int_{0}^{sqrt{1-z^2}}dxdydz=int_{0}^{1}2sqrt{1-z^2}dz$



      And then I would use substitution, namely, $z = sin{x}Rightarrow dz =cos{x}dx$, so $int_{0}^{1}2sqrt{1-z^2}dz=int_{0}^{frac{pi}{2}}2sqrt{1-sin{x}^2}cos{x}dx=2int_{0}^{frac{pi}{2}}cos^{2}{x}dx=2int_{0}^{frac{pi}{2}}frac{1}{2}+frac{1}{2}cos{2x}dx=int_{0}^{frac{pi}{2}}1+cos{2x}dx=x+frac{sin{2x}}{2}vert_{0}^{frac{pi}{2}}=frac{pi}{2}$



      So getting back to eight octants, we'd get $lambda^{d}(A)=4pi$ but I have been told this is incorrect. I do not understand where I went wrong






      real-analysis integration measure-theory




      real-analysis integration measure-theory






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      asked Jan 19 at 20:14









      SABOYSABOY

      656311




      656311






















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          $begingroup$

          $4pi$ is quite clearly too much: we should expect a volume close to (and a bit larger than) the volume of a unit sphere, i.e. $frac{4}{3}pi$. Assume that the value of $zin[-1,1]$ has been fixed. The $z$-section of our body is shaped as
          $$ left{begin{array}{rcl}x^2+y^2&leq& 1\x^2,y^2&leq &1-z^2end{array}right.$$
          hence it is a square with side length $2sqrt{1-z^2}$ for any $|z|geq frac{1}{sqrt{2}}$ and a circle with four circle segments being removed for any $|z|leqfrac{1}{sqrt{2}}$. In the former case the area of the section is $4(1-z^2)$, in the latter it is $pi + 4|z|sqrt{1-z^2}-4arcsin(|z|)$. The wanted volume is so



          $$ 2left[int_{0}^{1/sqrt{2}}4(1-z^2),dz + int_{1/sqrt{2}}^{1}pi + 4zsqrt{1-z^2}-4arcsin(z),dzright]$$
          i.e. $color{blue}{8sqrt{2}-2pi}approx 5$.






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            1 Answer
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            1 Answer
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            active

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            $begingroup$

            $4pi$ is quite clearly too much: we should expect a volume close to (and a bit larger than) the volume of a unit sphere, i.e. $frac{4}{3}pi$. Assume that the value of $zin[-1,1]$ has been fixed. The $z$-section of our body is shaped as
            $$ left{begin{array}{rcl}x^2+y^2&leq& 1\x^2,y^2&leq &1-z^2end{array}right.$$
            hence it is a square with side length $2sqrt{1-z^2}$ for any $|z|geq frac{1}{sqrt{2}}$ and a circle with four circle segments being removed for any $|z|leqfrac{1}{sqrt{2}}$. In the former case the area of the section is $4(1-z^2)$, in the latter it is $pi + 4|z|sqrt{1-z^2}-4arcsin(|z|)$. The wanted volume is so



            $$ 2left[int_{0}^{1/sqrt{2}}4(1-z^2),dz + int_{1/sqrt{2}}^{1}pi + 4zsqrt{1-z^2}-4arcsin(z),dzright]$$
            i.e. $color{blue}{8sqrt{2}-2pi}approx 5$.






            share|cite|improve this answer









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              0












              $begingroup$

              $4pi$ is quite clearly too much: we should expect a volume close to (and a bit larger than) the volume of a unit sphere, i.e. $frac{4}{3}pi$. Assume that the value of $zin[-1,1]$ has been fixed. The $z$-section of our body is shaped as
              $$ left{begin{array}{rcl}x^2+y^2&leq& 1\x^2,y^2&leq &1-z^2end{array}right.$$
              hence it is a square with side length $2sqrt{1-z^2}$ for any $|z|geq frac{1}{sqrt{2}}$ and a circle with four circle segments being removed for any $|z|leqfrac{1}{sqrt{2}}$. In the former case the area of the section is $4(1-z^2)$, in the latter it is $pi + 4|z|sqrt{1-z^2}-4arcsin(|z|)$. The wanted volume is so



              $$ 2left[int_{0}^{1/sqrt{2}}4(1-z^2),dz + int_{1/sqrt{2}}^{1}pi + 4zsqrt{1-z^2}-4arcsin(z),dzright]$$
              i.e. $color{blue}{8sqrt{2}-2pi}approx 5$.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                $4pi$ is quite clearly too much: we should expect a volume close to (and a bit larger than) the volume of a unit sphere, i.e. $frac{4}{3}pi$. Assume that the value of $zin[-1,1]$ has been fixed. The $z$-section of our body is shaped as
                $$ left{begin{array}{rcl}x^2+y^2&leq& 1\x^2,y^2&leq &1-z^2end{array}right.$$
                hence it is a square with side length $2sqrt{1-z^2}$ for any $|z|geq frac{1}{sqrt{2}}$ and a circle with four circle segments being removed for any $|z|leqfrac{1}{sqrt{2}}$. In the former case the area of the section is $4(1-z^2)$, in the latter it is $pi + 4|z|sqrt{1-z^2}-4arcsin(|z|)$. The wanted volume is so



                $$ 2left[int_{0}^{1/sqrt{2}}4(1-z^2),dz + int_{1/sqrt{2}}^{1}pi + 4zsqrt{1-z^2}-4arcsin(z),dzright]$$
                i.e. $color{blue}{8sqrt{2}-2pi}approx 5$.






                share|cite|improve this answer









                $endgroup$



                $4pi$ is quite clearly too much: we should expect a volume close to (and a bit larger than) the volume of a unit sphere, i.e. $frac{4}{3}pi$. Assume that the value of $zin[-1,1]$ has been fixed. The $z$-section of our body is shaped as
                $$ left{begin{array}{rcl}x^2+y^2&leq& 1\x^2,y^2&leq &1-z^2end{array}right.$$
                hence it is a square with side length $2sqrt{1-z^2}$ for any $|z|geq frac{1}{sqrt{2}}$ and a circle with four circle segments being removed for any $|z|leqfrac{1}{sqrt{2}}$. In the former case the area of the section is $4(1-z^2)$, in the latter it is $pi + 4|z|sqrt{1-z^2}-4arcsin(|z|)$. The wanted volume is so



                $$ 2left[int_{0}^{1/sqrt{2}}4(1-z^2),dz + int_{1/sqrt{2}}^{1}pi + 4zsqrt{1-z^2}-4arcsin(z),dzright]$$
                i.e. $color{blue}{8sqrt{2}-2pi}approx 5$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 19 at 22:20









                Jack D'AurizioJack D'Aurizio

                290k33283664




                290k33283664






























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