Finding volume of sphere cartesian to polarUsing cartesian coordinates the volume of a sphere can be calculated via ∫ − R R ∫ − R 2 − z 2 R 2 − z 2 ∫ − R 2 − z 2 − y 2 R 2 − z 2 − y 2 1 d x d y d z However using a parametrisation with polar coordinates the Integral becomes: ∫ 0 2 π ∫ 0 R ∫ 0 π 1 R 2 sin ⁡ ( θ ) d θ d R d φ .Thinking closer about this I just don't see how these integrals are interchangeable. Notably substituting polar coordinates in the cartesian one I don't notice how the borders simplify.For instance using the substitution ( x y z ) = ( R cos ⁡ ( φ ) sin ⁡ ( θ ) R sin ⁡ ( φ ) sin ⁡ ( θ ) R cos ⁡ ( θ ) ) I get for the first border: R 2 ( 1 − cos 2 ⁡ ( θ ) − sin 2 ⁡ ( θ ) sin 2 ⁡ ( φ ) ) . What's that even?

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Finding volume of sphere cartesian to polarUsing cartesian coordinates the volume of a sphere can be calculated via      ∫          −      R              R            ∫          −                        R          2                −                  z          2                                              R          2                −                  z          2                          ∫          −                        R          2                −                  z          2                −                  y          2                                              R          2                −                  z          2                −                  y          2                      1        d    x        d    y        d    z  However using a parametrisation with polar coordinates the Integral becomes:      ∫          0              2            π            ∫          0              R            ∫          0              π        1        R    2      sin  ⁡  (  θ  )        d    θ          d    R        d    φ  .Thinking closer about this I just don&#039;t see how these integrals are interchangeable. Notably substituting polar coordinates in the cartesian one I don&#039;t notice how the borders simplify.For instance using the substitution      (                            x                                      y                                      z                      )    =      (                            R                    cos          ⁡          (          φ          )                    sin          ⁡          (          θ          )                                      R                    sin          ⁡          (          φ          )                    sin          ⁡          (          θ          )                                      R                    cos          ⁡          (          θ          )                      )  I get for the first border:            R      2        (    1    −          cos      2        ⁡    (    θ    )    −          sin      2        ⁡    (    θ    )              sin      2        ⁡    (    φ    )    )  . What&#039;s that even?

selixaak · Accepted Answer

Step 1Such as   x  =  −            R      2        −          z      2        −          y      2            ∫          −      R              R            ∫          −                        R          2                −                  z          2                                              R          2                −                  z          2                          ∫          −                        R          2                −                  z          2                −                  y          2                                              R          2                −                  z          2                −                  y          2                      1        d    x        d    y        d    z  so we can also do  2      ∫          0              π            ∫          0              2      π            ∫          0              R            ρ    2    s  e  n  φ        d    ρ        d    θ        d    φ  with   x  =  ρ  s  e  n  (  φ  )  c  o  s  (  θ  )  y  =  ρ  s  e  n  (  φ  )  s  e  n  (  θ  )  z  =  ρ  c  o  s  (  φ  )Step 2Then       x    2    +      y    2    +      z    2    =      ρ    2    s  e      n    2    (  φ  )  (  c  o      s    2    (  θ  )  +  s  e      n    2    (  θ  )  )  +      ρ    2    c  o      s    2    (  φ  )  =      ρ    2    s  e      n    2    (  φ  )  +      ρ    2    c  o      s    2    (  φ  )  =      ρ    2

Lisantiom · Accepted Answer

Explanation:If   x  =  R  cos  ⁡  (  φ  )  sin  ⁡  (  θ  ),   y  =  R  sin  ⁡  (  φ  )  sin  ⁡  (  θ  ), and   z  =  R  cos  ⁡  (  θ  ), then                              x          2                +                  y          2                +                  z          2                                    =                  R          2                          cos          2                ⁡        (        φ        )                  sin          2                ⁡        (        θ        )        +                  R          2                          sin          2                ⁡        (        φ        )                  sin          2                ⁡        (        θ        )        +                  R          2                          cos          2                ⁡        (        θ        )                                          =                  R          2                                      (                                    sin          2                ⁡        (        θ        )        +                  cos          2                ⁡        (        θ        )                              )                                                            =                  R          2                .

Finding volume of sphere cartesian to polar

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