Triple integration ∭ x 2 + y 2 + z 2 ≤ 2 z x 2 y 2 d x d y d z

Question

Triple integration       ∭                  x        2            +              y        2            +              z        2            ≤      2      z            x    2        y    2    d  x  d  y  d  z

Blaine Foster · Accepted Answer

It&#039;s better to use spherical coordinatesThe region is       x    2    +      y    2    +      z    2    ≤  2  z , i.e.,       x    2    +      y    2    +  (  z  −  1      )    2    ≤  1So we can define the spherical coordinates   r  ,  θ  ,  ϕ such that  x  =  r  sin  ⁡  θ  cos  ⁡  ϕ  y  =  r  sin  ⁡  θ  sin  ⁡  ϕ  z  =  1  +  r  cos  ⁡  θAnd the integral becomes  I  =            ∫              r        =        0            1              ∫              θ        =        0            π              ∫              ϕ        =        0                    2        π                    r      6              sin      5        ⁡    θ          cos      2        ⁡    ϕ          sin      2        ⁡    ϕ    d    ϕ    d    θ    d    r  Integrating with respect r and with respect to   ϕ is straight forward, and the integral reduces to  I  =            π      28                  ∫              θ        =        0            π        s    i          n      5        θ    d    θ  The integral with respect to   θ is solved using the substitution   u  =  cos  ⁡  θ, then      ∫          sin      5        ⁡    θ    d    θ    =    −    ∫    (    1    −          u      2              )      2        d    u    =    −    (    u    −                  2        3                    u      3        +                            u          5                5              )  Hence,  I  =                    4        π            105

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