Is there a nice way to integrate this? ∭ x 2 + y 2 + z 2 ⩽ 1 x 2 n + y 2 n + z 2 n d V

Question

Is there a nice way to integrate this?      ∭                  x        2            +              y        2            +              z        2            ⩽      1            x          2      n        +      y          2      n        +      z          2      n            d    V

ipa1rafd · Accepted Answer

First note that for symmetry reasons, your integral is equal to  3      ∭                  x        2            +              y        2            +              z        2            ≤      1            z          2      n        d  x  d  y  d  z  .Now use spherical coordinates, which describe the domain of integration very well:  =  3      ∫          0      ≤      r      ≤      1            ∫          0      ≤      ϕ      ≤      π            ∫          0      ≤      θ      ≤      2      π        (  r  cos  ⁡  ϕ      )          2      n            r    2    sin  ⁡  ϕ  d  r  d  ϕ  θ  .  =  3      (          ∫              0            1              r              2        n        +        2              d    r    )        (          ∫              0            π        (    cos    ⁡    ϕ          )              2        n              sin    ⁡    ϕ    d    ϕ    )        (          ∫              0                    2        π              1    d    θ    )    =  3      1          2      n      +      3            2          2      n      +      1        2  π  =            12      π              (      2      n      +      1      )      (      2      n      +      3      )        .

Chloe Arnold · Accepted Answer

Another way:                                            ∫                                    x              2                        +                          y              2                        +                          z              2                        ≤            1                                    x                      2            n                          +                  y                      2            n                          +                  z                      2            n                          d        x        d        y        d        z                            =                    3                  ∫                                    x              2                        +                          y              2                        +                          z              2                        ≤            1                                    x                      2            n                          d        x        d        y        d        z                            =                    3                  ∫                      −            1                                1                          d        x                  (                      x                          2              n                                            ∫                                          y                2                            +                              z                2                            ≤              1              −                              x                2                                              d          y          d          z          )                                    =                    3        π                  ∫                      −            1                                1                          d        x                  (                      x                          2              n                                (          1          −                      x            2                    )          )                                    =                    3        π                  (                      2                          2              n              +              1                                −                      2                          2              n              +              3                                )                                    =                                          12            π                                (            2            n            +            1            )            (            2            n            +            3            )

Is there a nice way to integrate this? int int int_(x^2+y^2+z^2<=1) x^(2n)+y^(2n)+z^(2n)dV

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