Compute the volume enclosed by the surface ( x 2 a 2 + y 2 b 2 + z 2 c 2 ) 2 = x 2 a 2 + y 2 b 2 − z 2 c 2 .

Question

Compute the volume enclosed by the surface            (                        x          2                          a          2                    +                        y          2                          b          2                    +                        z          2                          c          2                    )        2    =            x      2              a      2        +            y      2              b      2        −            z      2              c      2        .

Juliet Mcdonald · Accepted Answer

cos  ⁡      (    π    −          1      2              cos              −        1              ⁡    (    −          ρ      2        )    )    =If   0  ≤      ρ    2    ≤  −  cos  ⁡  (  2  θ  ), then             (      4      n      −      3      )      π        4    ≤  θ  ≤            (      4      n      −      1      )      π        4     for   n  ∈      Z  for             (      4      n      −      3      )      π        4    ≤  θ  ≤            (      4      n      −      1      )      π        4     for   n  ∈      Z  The polar angle must generally be between 0 and   π, so we take   θ  ∈      [          π      4        ,                  3        π            4        ]  . Then the volume is      ∫    0          2      π            ∫                  π        4                                      3          π                4                  ∫    0                  −        cos        ⁡        (        2        θ        )              a  b  c      ρ    2    sin  ⁡  (  θ  )    d  ρ    d  θ    d  φ  =            a      b      c              π        2                    4              2            Or, preserving the order you chose, we have      ρ    2    =  −  cos  ⁡  (  2  θ  )    ⟹    θ  =      1    2        cos          −      1        ⁡  (  −      ρ    2    )but this is only true if   θ  ∈      [    0    ,          π      2        ]  . For   θ  ∈      [          π      2        ,    π    ]  , we instead would have      ρ    2    =  −  cos  ⁡  (  2  θ  )    ⟹    θ  =  π  −      1    2        cos          −      1        ⁡  (  −      ρ    2    )Now, if   θ  ≤      π    2  , then  ρ  =  0    ⟹    θ  =            3      π        4     and   ρ  =  1    ⟹    θ  =      π    2  Otherwise, if       π    2    &amp;lt;  θ, then  ρ  =  0    ⟹    θ  =            3      π        4     and   ρ  =  1    ⟹    θ  =      π    2  All this tells us that if   ρ  ∈  [  0  ,  1  ], then   θ  ∈      [          1      2              cos              −        1              ⁡    (    −          ρ      2        )    ,    π    −          1      2              cos              −        1              ⁡    (    −          ρ      2        )    ]  The subsequent integral agrees with the previous result:                                            ∫          0                      2            π                                    ∫          0          1                          ∫                                    1              2                                      cos                              −                1                                      ⁡            (            −                          ρ              2                        )                                π            −                          1              2                                      cos                              −                1                                      ⁡            (            −                          ρ              2                        )                          a        b        c                  ρ          2                sin        ⁡        (        θ        )                d        θ                d        ρ                d        φ                                          =        2        a        b        c        π                  ∫          0          1                          ρ          2                          (          cos          ⁡                      (                          1              2                                      cos                              −                1                                      ⁡            (            −                          ρ              2                        )            )                    −          cos          ⁡                      (            π            −                          1              2                                      cos                              −                1                                      ⁡            (            −                          ρ              2                        )            )                    )                        d        ρ                                          =        4        a        b        c        π                  ∫          0          1                          ρ          2                cos        ⁡                  (                      1            2                                cos                          −              1                                ⁡          (          −                      ρ            2                    )          )                        d        ρ                                          =        2                  2                        a        b        c        π                  ∫          0          1                          ρ          2                          1          −                      ρ            2                                  d        ρ                                          =        2                  2                        a        b        c        π        ×                  π          16                =                              a            b            c                          π              2                                            4                          2

Compute the volume enclosed by the surface ( x 2

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