the lower limit is 0 the upper limit is x on the first and thirdintegral. the second integral lower limit is 0 andupper limit is 2x. ∫ sin ⁡ θ d θ ∫ d ψ ∫ r 2 d r

Question

the lower limit is 0 the upper limit is x on the first and thirdintegral.   the second integral lower limit is 0 andupper limit is 2x.  ∫  sin  ⁡  θ  d  θ  ∫  d  ψ  ∫      r          2            ﻿    d  r

Jazlene Dickson · Accepted Answer

∫          0              x            ∫          0              2      x            ∫          0              x        sin  ⁡  θ      r          2        d  r  d  ψ  d  θ  =      ∫          0              x            ∫          0              2      x            1    3        x          3        sin  ⁡  θ  d  ψ  d  θ  =      ∫          0              x        (      2    3        x          4        )  sin  ⁡  θ  d  θ  =      2    3        x          4        [  −  cos  ⁡  θ  ]            |              0              x        =            2              x                  4                      3    [  −  cos  ⁡  x  +  cos  ⁡  (  0  )  ]  =            2              x                  4                      3    [  1  −  cos  ⁡  (  x  )  ]

vangstosiis · Accepted Answer

∫  sin  ⁡  θ  d  θ  ∫  d  ψ  ∫      r          2        d  r  =      ∫          0              2            ∫          0              2      x        ∫      r          2        d  r  d  ψ  d  θ  =            2              r                  3                            x                  2                      3

the lower limit is 0 the upper limit is x on the first and thirdintegral. the second integral lower limit is 0 andupper limit is 2x. int sin theta dtheta int d psi int r^2 dr=?

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