a) Show that the formula for the surface area of a sphere with radius r is 4 π r 2 .b) If a portion of the sphere is removed to form a spherical cap of height h then then show the curved surface area is 2 π h r 2 ?

Question

a) Show that the formula for the surface area of a sphere with radius r is       4    π          r      2      .b) If a portion of the sphere is removed to form a spherical cap of height h then then show the curved surface area is       2    π    h          r      2      ?

Emencisydeessyo2 · Accepted Answer

A sphere&#039;s area element has a constant radius r and two angles. One is longitude       ϕ  , which varies from 0 to       2    π  . The other one is the angle with the vertical. To avoid counting twice, that angle only varies between 0 and       π  .Thus, the area element is       d    A    =    r    d    θ    r          sin      θ        d    ϕ    =          r      2              sin      θ        d    θ    d    ϕ  Integrated over the whole sphere gives      A    =                  ∫        0        π                    sin      θ        d    θ                  ∫        0                  2          π                      d    ϕ          r      2        =    -          cos              (        θ        )                                      ∣                0                  π                      2    π          r      2        =    4    π          r      2      In part b,             cos              (        θ        )             varies between             a      r       and             b      r       which is such that       b    -    a    =    h  Thus       A    =          cos      θ                                ∣                b        a              2    π          r      2        =          (              b        r            -              a        r            )        2    π          r      2        =          (              h        r            )        2    π          r      2        =    2    π    r    h  Every other derivation of this result that I found uses cylindrical coordinates and is far more involved than this one.

smakkie8iz · Accepted Answer

Spherical coordinates are more convenient to use than cylindrical or rectangular coordinates. This solution appears lengthy because I have broken down each step, but it can be computed in a few lines of code.With spherical coordinates, we can define a sphere of radius r by all coordinate points where       0    ≤    ϕ    ≤    π  (Where       ϕ   is the angle measured down from the positive z-axis), and       0    ≤    θ    ≤    2    π   (just the same as it would be polar coordinates), and       ρ    =    r  ). The Jacobian for Spherical Coordinates is given by       J    =          ρ      2              sin      ϕ      And so we can calculate the surface area of  a sphere of radius r using a double integral:      A    =    ∫          ∫      R                  d    S            where       R    =          {              (        x        ,        y        ,        z        )            ∈              ℝ        3                    ∣                    x        2            +              y        2            +              z        2            =              r        2            }            ∴    A    =                  ∫        0        π                                 ∫        0                  2          π                                 r      2              sin      ϕ             d    θ         d    ϕ  When we look at the inner integral, we get:                    ∫        0                  2          π                                 r      2              sin      ϕ             d    θ    =          r      2              sin      ϕ                           ∫        0                  2          π                           d    θ                                                       =          r      2              sin      ϕ                                [                     θ                     ]                0                  2          π                                                                         =          (              r        2                    sin        ϕ            )              (      2      π      -      0      )                                                           =    2    π          r      2              sin      ϕ      Thus, our integral becomes:      A    =                  ∫        0        π                   2    π          r      2              sin      ϕ             d    ϕ                       =    -    2    π          r      2                                {                      cos            ϕ                    ]                0        π                                 =    -    2    π          r      2              (              cos        π            -              cos        0            )                           =    -    2    π          r      2              (      -      1      -      1      )                           =    -    2    π          r      2              (      -      2      )                           =    4    π          r      2                 QEDBy trigonometry             cos      ϕ        =          h      r        ⇒    ϕ    =          arccos              (                  h          r                )            , and so we must restrict       ϕ   to             arccos              (                  h          r                )              ≤    ϕ    ≤          π      2      , which gives us:      A    =                  ∫                  arccos                      (                          h              r                        )                                                π            2                                                   ∫        0                  2          π                                 r      2              sin      ϕ             d    θ         d    ϕ                       =                  ∫                  arccos                      (                          h              r                        )                                                π            2                                           (              r        2                    sin        ϕ            )              (      2      π      -      0      )             d    ϕ                       =                  ∫                  arccos                      (                          h              r                        )                                                π            2                                     2    π          r      2              sin      ϕ             d    ϕ                       =    -    2    π          r      2                                [                      cos            ϕ                    ]                          arccos                      (                          h              r                        )                                                π            2                                                   =    -    2    π          r      2              (              cos                  (                      π            2                    )                    -              cos                  (                      arccos                          (                              h                r                            )                                )                    )                           =    -    2    π          r      2              (      0      -              h        r            )                           =    -    2    π          r      2              (      0      -              h        r            )                           =    2    π    h          r      2                 QED

Show that the formula for the surface area of a sphere with radius r is 4pi r^2. b) If a portion of the sphere is removed to form a spherical cap of height h then then show the curved surface area is 2 pi h r^2?

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