Finding a volume integral in an ellipsoidI am trying to find the volume integral of ρ = ρ 0 ( R 2 − r 2 R 2 ) inside an ellipsoid given by x 2 ( 3 R ) 2 + y 2 ( 4 R ) 2 + z 2 ( 5 R ) 2 = 1I've tried using jacobian to move from an ellipsoid to an unit ball by these mapping relationships x = 3 R u, y = 4 R v, z = 5 R w.But the resulting integral is still heavy ∫ ρ 0 ( 1 − ( 9 u 2 + 16 w 2 + 25 w 2 ) ) 60 d u d v d wDoes anyone have any insight to a more elegant way.

Question

Finding a volume integral in an ellipsoidI am trying to find the volume integral of   ρ  =      ρ          0            (                            R                      2                          −                  r                      2                                      R                  2                      )   inside an ellipsoid given by             x              2                    (      3      R              )                  2                      +            y              2                    (      4      R              )                  2                      +            z              2                    (      5      R              )                  2                      =  1I&#039;ve tried using jacobian to move from an ellipsoid to an unit ball by these mapping relationships   x  =  3  R  u,   y  =  4  R  v,   z  =  5  R  w.But the resulting integral is still heavy  ∫      ρ    0        (    1    −          (      9              u                  2                    +      16              w                  2                    +      25              w                  2                    )        )    60  d  u  d  v  d  wDoes anyone have any insight to a more elegant way.

wietselau · Accepted Answer

Explanation:Your last integral is not that heavy.The constant 1 yields the volume of the unit sphere. Then using spherical coordinates, the integral of       w    2   is      ∫    0          2      π            ∫    0    π        ∫    0    1        r    2        cos    2    ⁡  ϕ        r    2    sin  ⁡  ϕ    d  r    d  ϕ    d  θ  =  2  π      2    3        1    5    ..By symmetry, the integrals of       u    2   and       v    2   are equal.

Massatfy · Accepted Answer

Step 1The resulting integral in this case is actually very simple due to symmetry:                              ∫          B                          u          2                        d        V        =                  ∫          B                          v          2                        d        V        =                  ∫          B                          w          2                        d        V        =                  1          3                          ∫          B                (                  u          2                +                  v          2                +                  w          2                )                d        V        =                  1          3                          ∫          0          1                          r          2                        4        π                  r          2                        d        r        =                              4            π                    15                .            Step 2This symmetry should be intuitive enough. Otherwise use the change of variables theorem to permute (u, v, w) and the fact that the unit ball remains invariant under this change of coordinates, and that the absolute value of the determinant of this is 1 if you want a really rigorous proof.

Find the volume integral of rho=rho_{0}(R^2-r^2/R^2) inside an ellipsoid given by x^2/((3R)^2)+y^2/((4R)^2)+z^2/((5R)^2)=1

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