ENGI 5432/5435 Advanced Calculus
Final Examination, 2008 — Solutions

Note: You need the Symbol font (68 kB) in order to display various symbols correctly, such as (otherwise they appear as p, r, q and f ).

Find the Fourier series expansion on the interval (–1, 1) of the function

The function f (x) is odd

$b_n = 2*Integral_0^1 { (x^3 - x) sin(n pi x)} dx$
(using the symmetry property for integrals of odd functions)

Therefore the Fourier series for f (x) on (–1, 1) is

$f(x) = 12/pi^3 Sum{(-1)^n sin(n pi x) / n^3}$

The plot below illustrates how good an approximation the Fourier series is, even after the partial sum of only the first two terms:

The vector field is defined by .
1. Show that everywhere in ³.
2. Hence evaluate , where S is the surface of the sphere of radius 1 and centre at (0,0,0). State the name of the theorem that you are using.
  [You may express your answer in terms of p.]
  
  By Gauss’ divergence theorem,
  
  where V is the volume enclosed by S .
  But the volume of a sphere of radius 1 is
  .
  Therefore the flux through the sphere is

In spherical polar coordinates (r, q, f), a scalar function V is defined on the domain
W = { all of ³ except the z-axis } by

V(r, q, f) = ln (r sin q )
1. Find in spherical polar coordinates.
  
  Therefore, in spherical polar coordinates,
2. Express V as a function of the Cartesian coordinates (x, y, z) only.
  
  $V = ln sqrt{x^2 + y^2}$
3. Use your answer to part (b) to find in Cartesian coordinates.
  
  $grad V = < x, y, 0 > / sqrt{x^2 + y^2}$
4. Show that your answers to parts (a) and (c) are equivalent by using the appropriate coordinate conversion matrix.
  
  Either
  
  First note that
  
  Converting the answer to part (c) into spherical polar coordinates,
  
  which matches the answer to part (a)
  
  or
  
  Converting the answer to part (a) into Cartesian coordinates,
  
  which matches the answer to part (c)
5. Find the Laplacian in one of the two coordinate systems.
  
  Either in spherical polar coordinates
  
  (except possibly on the z-axis, where r sin q = 0). Therefore on W
  
  or in Cartesian coordinates
  
  (except possibly on the z-axis, where x² + y² = 0). Therefore on W
6. Is V harmonic on any domain that excludes the z axis?
  
  YES

For the partial differential equation
1. Classify this partial differential equation (as one of hyperbolic, parabolic or elliptic).
  
  Therefore the PDE is
  
  hyperbolic
2. Find the general solution u(x, y).
  
  The complementary function is
  
  (where f and g are arbitrary functions).
  The right side is a zero order polynomial (a constant)
  For the particular solution, try second order polynomials, without the lower order terms:
  
  Substitute this trial particular solution into the PDE:
  
  There is only one constraint on three unknowns, leaving two free choices.
  Choose a = b = 0, then c = 1.
  [Other choices are valid, but this is the easiest.]
  The particular solution is u_P = y² and the general solution is
  
  The choice b = 0 , a = c leads to the alternative form
3. Using the additional information
  
  Find the complete solution u(x, y).
  
  Following the choice made in part (b),
  
  [The choice 2(B) + (C) to find f(x) first is also valid.]
  
  Substituting g(–2x) back into equation (A):
  
  [Note that the arbitrary constants of integration for the two functions f(x) and g(x) cancel each other out.]
  The general solution becomes the complete solution
  
  The complete solution is therefore
  
  [It is easy to verify that this solution does satisfy the PDE and both conditions.]

Find the equations of the line of force for the vector field that passes through the point (1, 1, 1).

Therefore the family of lines of force is

[The solution is part of this general solution.]
The required line of force passes through (1, 1, 1).

Therefore the required line of force is

A vector is defined in the cylindrical polar coordinate system by .
1. Find the derivative in terms of .
  
  Therefore
2. Show whether or not a potential function exists for and, if it does exist, on what domain.
  
  Therefore
  
  a potential function does not exist anywhere in ³.

Find the value of the line integral

$W = integ_C { 2(x+y) dx + (2x+2y-1) dy }$

where C is the arc of the parabola y = x² between (–1, 1) and (1, 1).

Check for any potential function:

A potential function therefore exists and the value of the line integral is simply the potential difference between the two endpoints.

Therefore

OR

The line integral can be evaluated directly. Along the parabolic path the natural parameters are

$W = Integral { 6t^2 + 4t^3} dt$

Note that the inverse of y = x² is along the right half of the path (x=0 to x=1), but is along the left half of the path (x=–1 to x=0).
Another direct evaluation of the line integral is therefore

A shell is in the shape of that part of the ellipsoid that is above the x-y plane (z > 0).
Its surface density is $rho = 36z / sqrt{16x^2 + 81y^2 + 1296z^2}$
Find the mass m of this shell.
Note: For the upper half of the general ellipsoid , a parametric net is (q, f), such that .
For the projection method, start with $z = sqrt{ 1 - x^2 / 9 - y^2 / 4 }$ .
If necessary, you may quote $Integral of sqrt{a^2 - x^2} = x sqrt{a^2 - x^2} / 2 + a^2 Arcsin(x/a) / 2 + C$ .
Either method (parametric net or projection) may be used in this question.

Parametric Net Method:

For this ellipsoid, a = 3, b = 2, c = 1

The outward normal vector at every point on the ellipsoid is

Converting the density function to the new coordinate system:

The mass of the shell is

Therefore the mass of the shell is

OR

Projection Method:

A normal vector to the surface is

$scalar N = sqrt{16x^2 + 81y^2 + 1296z^2} / (36z)$

The mass of the shell is then

where A is the area of the shadow of the ellipsoid on the x-y plane, that is,
the area of an ellipse of semi-major axis a = 3 and semi-minor axis b = 2.

Therefore the mass of the shell is

In the event that the formula for the area of an ellipse is not immediately available, the area can be calculated in Cartesian coordinates as follows.

The area of the entire ellipse is four times the area of that part of the ellipse in the first quadrant.
$Area of ellipse = 4 Integral { sqrt(1 - x^2 / 9) }$
Using the integration identity in the question,
$Integral of sqrt{9 - x^2} = [x sqrt{9 - x^2} / 2 + 9 Arcsin(x/3) / 2] = 6 pi$ .