## 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 ³ ¹/6 from the z-axis, as seen from the origin.

a) sketch this region

In[12]:=

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

Out[12]=

```  -Graphics3D-
```

In[13]:=

```  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 -> {-4,4}]
```

Out[13]=

```  -Graphics3D-
```

In[14]:=

`  Show[{pic1,pic2}]`

Out[14]=

```  -Graphics3D-
```

In the spherical coordinate system, you would have

In[15]:=

```  Integrate[rho^2 Sin[phi]^2 Sin[theta]^2  * rho^2 Sin[phi], {phi,Pi/6, Pi}
\
{rho, 0, 4},
{theta, 0,2 Pi}]```

The extras after the * are because the differential is rho^2 drho sin(phi)
dtheta dphi. 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[16]:=

`  Integrate[rho^2 Sin[phi]^2 Sin[theta]^2  * rho^2 Sin[phi],{rho,0,4}]`

Out[16]=

```               3           2
1024 Sin[phi]  Sin[theta]
--------------------------
5
```

In[17]:=

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

Out[17]=

```                  3
1024 Pi Sin[phi]
-----------------
5
```

In[18]:=

`  Integrate[%, {phi, Pi/6, Pi}]`

Out[18]=

```  2048 Pi   384 Sqrt[3] Pi
------- + --------------
15            5
```

In[19]:=

`  N[%]`

Out[19]=

```  846.831
```

Up to Test 3 solutions