Problem Set 6   -   Solutions

---

Find the centroid (centre of mass when the surface density is constant) of the frustum of the circular cone   z² = x² + y²   between the origin and the plane   z = a, (where   a   is a positive constant).

Surface density   r = constant Use parameters (q, z) (0 £ q £ 2p,     0 £ z £ a )

Taking moments about the   x-y plane:
DM_z = z Dm    
   
 

Therefore the centroid of the frustum of the cone is at  

Find the total flux that passes through the hemisphere   x² + y² + z² = a²,   z > 0 ,   due to the electric flux density     using:

1. the surface projection method   and

   ---

   S:     x ² + y ² + z ² = a ²,     z > 0  
   
       (by symmetry)
    
   
   But everywhere on the hemisphere   r = a
   
   Switch to plane polar coordinates, in which   dA = r dr df.
   x = r cos f,     y = r sin f,     Þ     x ² + y ² = r ²

   

     everywhere on the hemisphere
    
   

   

---

2. the parametric surface net method.

   ---

   Use the parameters   (q, f)       (0 £ q £ p/2,     0 £ f < 2p).
   

   

   
   [Note:   this vector   N   has a different magnitude from the vector   N   in part (a) of this question.]
   
   
   
   
   But everywhere on the hemisphere   r = a  
   
   
      

---

3. What can you say about the relative orientation of the vector field D and the normal N to the hemisphere?

   ---

   From part (b) above, vectors   D   and   N   are both multiples of the displacement vector   r.   D   and   N   are therefore parallel to each other everywhere on the hemisphere.

Find the total flux of the vector field through that part of the circular paraboloid   z = 4 - x² - y²   that lies in the first octant.

Projection method:
z = 4 - x ² - y ²     (first octant only)


Use plane polar coordinates
      x = r cos f,     y = r sin f,       Þ     x ² + y ² = r ²
Þ     z = 4 - r ²     and



Therefore

Net flux through   S = p.

or

Surface parametric net method:
Use parameters   r   and   f:
      x = r cos f,     y = r sin f,     z = 4 - r ²
 

A thin sheet, in the shape of the paraboloid   z = 16 - (x² + y²), where   x > 0   and   y > 0, has a surface density of     It lies within the coaxial region between the circular cylinders     and  

1. Determine the total mass of the sheet, and

   ---

   In order to obtain a better visualization of the surface S, examine its projection onto any vertical plane containing the z-axis and its projection onto the x-y plane:

    

   Let   r ² = x ² + y ²,   then the paraboloid is
         z = 16 - r ²
   Cylinder   C₁   is     r ² = 1
   Cylinder   C₂   is     r ² = 9
   The sheet is therefore defined by
         z = 16 - r ²,     7 £ z £ 15,     x ³ 0 ,     y ³ 0.

   Surface parametric net method:

   Use parameters   (r, f):
   
   1 £ r £ 3 ,     0 £ f £ p/2
   The surface density then becomes
   
   
   
    
   
   Total mass:
   
   
   Therefore (assuming S.I. units)

   Total mass = 10 kg

   Projection Method:

   z = 16 - (x ² + y ²)
   
   
   
   
   Convert to plane polar coordinates:
   
   
   Therefore (assuming S.I. units)

   Total mass = 10 kg

---

2. Locate the centre of mass for the sheet.

   ---

   Elements of first moments:
   DM_x   =   x . Dm   =   x r DS   =   x r N Dr Df
   
   
   
   
   By symmetry,  

   DM_z   =   z . Dm   =   z r DS   =   z r N Dr Df
   
   
   
    
   
   
   Therefore the centre of mass is at

   

   Or, to 3 d.p.,

   (1.613,   1.613,   9.933)

Fluid is flowing along a cylindrical pipe.   The circular cross section inside the pipe has a constant radius of a (m).   The Cartesian coordinate system is aligned with the x axis along the line of symmetry (centre line) of the pipe.   The other two axes are in the plane of one of the circular cross sections.   The velocity of fluid at any point (x, y, z) inside the pipe is
             
Find the flux   Q (in m³/s)   (the rate at which fluid is flowing through the pipe).

The x-z and y-z planes are illustrated here, together with the velocity profile in the former case:

The speed increases from zero on the wall of the pipe to a maximum of   v ₀   along the centre line.
The flux   Q   is the flux of the velocity vector   v   of the fluid across a circular cross section in the direction of a unit normal to that cross section.   The unit normal is


Transform to plane polar coordinates:

[We can also verify that this answer is dimensionally consistent:
the units on the right side are [m s^–1] × [m]² = [m³ s^–1]
which are the units for volume per unit time.]

Complete the evaluation (in Example 2.6.3 of the lecture notes) of the line integral     around the unit square in the x-z plane for , (without using Stokes’ theorem).

From example 2.6.3, the positive orientation around the square is   OGHJ (the y axis is directed into the page).
In the x-z plane y = 0  
Compute the line integral around the four sides of the square, starting with OG:



The above was presented in a lecture.
Below are the calculations for the other three sides of the square:










Therefore

1. using a parametric net method. and

   ---

   Use a parametric net (q, f) that is a close relative of the spherical polar coordinate system:
   x = 2 sin q cos f ,       y = 2 sin q sin f ,       z = cos q
   Þ     x² + y²   =   4 sin²q
   and     x² + y² + 4z²   =   4 sin²q   + 4 cos²q   =   4
   The parameter ranges are     0 < q < p/2 ,     0 < f < 2p
   Employ the abbreviations     c_q = cos q ,   s_q = sin q ,   c_f = cos f ,   s_f = sin f
   
   
   
   
   
   
   Therefore

   

   By symmetry,  
   Taking moments about the xy-plane:
   
   
   
   Therefore the centre of mass is at the point

   

---

2. using a projection method.

   ---

   The shadow of the ellipsoid on the xy-plane is the area bounded by the circle   x² + y² = 4.
   Everywhere on the ellipsoid,  
   
   
   
   
   
   
   
   Switch to plane polar coordinates (r, q):
   
   Therefore

   

   By symmetry,  
   Taking moments about the xy-plane:
   
   
   But the projection is the region bounded by a circle of radius 2.
   Þ     A   =   p (2)²
   
   Therefore the centre of mass is at the point