3 (4 points). A solid S consists of the part of a ball of radius 4 that lies
within a cone with its vertex at the origin and consisting of points within
angle phi ² ¹/6 from the z-axis, as seen from the origin.

In[10]:=

  pic1 = ParametricPlot3D[{4 Sin[phi] Cos[theta], \
      4 Sin[phi] Sin[theta], 4 Cos[phi]},{phi,0,Pi/6},\
                {theta,0,2 Pi}, PlotRange -> {0,4}]
               

Out[10]=

  -Graphics3D-

In[11]:=

  pic2 = ParametricPlot3D[{rho Sin[Pi/6] Cos[theta], \
      rho Sin[Pi/6] Sin[theta], rho Cos[Pi/6]},{rho,0,4},\
                {theta,0,2 Pi}, PlotRange -> {0,4}]
               

Out[11]=

  -Graphics3D-

In[12]:=

  Show[{pic1,pic2}]

Out[12]=

  -Graphics3D-

In the cylindrical coordinate system, you would have

In[13]:=

  Integrate[r^2 Cos[theta]^2  * r, {theta,0, 2 Pi} \
        {r, 0, 4 Sin[Pi/6]}, 
        {z, r Cot[Pi/6],Sqrt[16-r^2]}]

The extra r is because the differential is r dr dtheta dz. Remember that the
order or operations in Mathematica is like the order of the integral signs
(last first). While Mathematica will execute the command just given, I find
it is often better to help her along:

In[14]:=

  Integrate[r^2 Cos[theta]^2  * r,{z, r Cot[Pi/6],Sqrt[16-r^2]}]

Out[14]=

                   4           2     3            2            2
      = -(Sqrt[3] r  Cos[theta] ) + r  Sqrt[16 - r ] Cos[theta]

In[15]:=

  Integrate[%, {r, 0, 4 Sin[Pi/6]}]

Out[15]=

                       2                         2
        2048 Cos[theta]    384 Sqrt[3] Cos[theta]
      = ---------------- - -----------------------
               15                     5

In[16]:=

  Integrate[%, {theta, 0, 2 Pi}]

Out[16]=

        128 (16 - 9 Sqrt[3]) Pi
      = -----------------------
                  15

In[17]:=

  N[%]

Out[17]=

  11.0327

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