Finding the volume of a cube using spherical coordinatesCalculate the volume of a cube having edge length a by integrating in spherical coordinates. Suppose that the cube have all the edges on the positive semi-axis. Let us divide it by the plane passing through the points (0;0;0),(0;0;a),(a;a;0).We get two equivalent prism; now we divide the section again by the plane passing through the points (a;0;a),(0;0;0),(a;a;a); So one may write: 1 2 V = ( ∫ 0 π / 4 d ϕ ∫ 0 π / 4 d θ ∫ 0 a cos ⁡ ϕ r 2 sin ⁡ ϕ d r + ∫ π / 4 π / 2 d ϕ ∫ 0 π / 4 d θ ∫ 0 a sin ⁡ ϕ cos ⁡ θ r 2 sin ⁡ ϕ d r )So one should expect V = a 3 but the latter expression gives a different result. What is the correct way to calculate V and why doesn't my reasoning work?

Question

Finding the volume of a cube using spherical coordinatesCalculate the volume of a cube having edge length a by integrating in spherical coordinates. Suppose that the cube have all the edges on the positive semi-axis. Let us divide it by the plane passing through the points (0;0;0),(0;0;a),(a;a;0).We get two equivalent prism; now we divide the section again by the plane passing through the points (a;0;a),(0;0;0),(a;a;a); So one may write:      1    2    V  =  (      ∫          0              π              /            4          d  ϕ      ∫          0              π              /            4          d  θ      ∫          0                      a                  cos          ⁡          ϕ                          r    2    sin  ⁡  ϕ    d  r    +        ∫          π              /            4              π              /            2          d  ϕ      ∫          0              π              /            4          d  θ      ∫          0                      a                  sin          ⁡          ϕ          cos          ⁡          θ                          r    2    sin  ⁡  ϕ    d  r  )So one should expect   V  =      a    3   but the latter expression gives a different result. What is the correct way to calculate V and why doesn&#039;t my reasoning work?

Alison Mcgrath · Accepted Answer

Step 1Your limits of the integration for   ϕ are wrong, because for a cube   ϕ  =  ϕ  (  θ  ) in your notation. The integrals with correct limits are                              1          2                V                            =                  ∫                      0                                π                          /                        4                                    ∫                      0                                arctan            ⁡            (                          a                              cos                ⁡                θ                                      )                                    ∫                      0                                              a                              cos                ⁡                ϕ                                                              r          2                sin        ⁡        ϕ                d        r                d        ϕ                d        θ                                          +                  ∫                      π                          /                        4                                π                          /                        2                                    ∫                      arctan            ⁡            (                          a                              cos                ⁡                θ                                      )                                π                          /                        2                                    ∫                      0                                              a                              sin                ⁡                ϕ                cos                ⁡                θ                                                              r          2                sin        ⁡        ϕ                d        r                d        ϕ                d        θ                                          =                              a            3                    6                +                              a            3                    3                =                              a            3                    2                    Step 2Solving the integrals is a bit tedious, but straightforward. For doing the integration it is helpful to note that   ∫            (              1                  cos          ⁡          (          ϕ          )                    )        2      d  ϕ  =  tan  ⁡  ϕ  +  C and   ∫            (              1                  cos          ⁡          (          ϕ          )                    )        3    sin  ⁡  ϕ    d  ϕ  =  ∫            (              1                  cos          ⁡          (          ϕ          )                    )        2    tan  ⁡  ϕ    d  ϕ  =      1    2    (  tan  ⁡  ϕ      )    2    +  C

Calculate the volume of a cube having edge length a by integrating in spherical coordinates.

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