]> Exercise 3

Math Animated™
Mathematical Introduction for Physics and Engineering
by Samuel Dagan (Copyright © 2007-2020)

Chapter 3: Many Variables; Section 2: Integration; page 3

Two Dimensional Transformation, Exercise 3

Question

Answer the following questions by use of the transformation of coordinates $(x, y) \Leftrightarrow (β, γ)$

$\begin{array}{l} x = γ a \cos β \\ y = γ b \sin β \end{array}$
where a and b are positive constants, and x, y, a, b have physical dimensions of length.

The density (mass/area) of the uniform planar solid ellipse

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
is σ . What is the moment of inertia I_y about the y axis?

As a continuation of question 1, two circular holes with radius ρ each contained fully by the ellipse are perforated. The centres of the holes are located at (x,y)=(±d,0), where d>ρ . What is the new moment of inertia? (Hint: Steiner's theorem)

Without relation to the questions 1 and 2, what is the volume enclosed between the partial surface of the elliptic paraboloid

$z = h (1 - \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}) \geq 0$
and the (x,y) plane, where h is a positive constant? Compare this with the results for the same volume done in a previous exercise by a different method.

Check the final results of questions 1,2,3 for consistency with physical dimensions!

Reminder

... The change of two variables ...

$\begin{array}{l} x = x (u, v) \\ y = y (u, v) \end{array}}$	(3.2.3.3)

..... transformation of coordinates ....

.... the magnification factor μ of a rectangular element of "area" ΔuΔv, mapped to the corresponding element ΔxΔy from the original plane:

$\begin{array}{l} F (x, y) Δ x Δ y = F [x (u, v), y (u, v)] μ (u, v) Δ u Δ v \\ or Δ x Δ y = μ (u, v) Δ u Δ v \end{array}}$	(3.2.3.5)

... the magnification factor μ ..... is the absolute value of the Jacobian :

$Δ x Δ y = \| \frac{D (x, y)}{D (u, v)} \| Δ u Δ v = \| \frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial y}{\partial u} \| Δ u Δ v$	(3.2.3.6)

and therefore the balance of the physical dimensions is kept.

As an example .... we can form a transformation of coordinates:

$\begin{array}{l} x = γ a \cos β \\ y = γ b \sin β \end{array}}$	(3.2.3.9)

where β and γ are the new variables.... For any ...γ, one obtains an ellipse with the same ratio of radii (a/b), for γ=1 the radii are a and b. Any point enclosed in this ellipse, is mapped to a point enclosed in a rectangle of the new coordinate system. The "area" in the (β,γ) coordinate system corresponds to:

$\int_{0}^{1} d γ \int_{0}^{2 π} d β = 2 π$	(3.2.3.10)

which does not make any sense, without using the Jacobian as the magnification factor, which is

$μ (β, γ) = \| \frac{\partial x}{\partial β} \frac{\partial y}{\partial γ} - \frac{\partial x}{\partial γ} \frac{\partial y}{\partial β} \| = γ a b \| - \sin^{2} β - \cos^{2} β \| = γ a b$	(3.2.3.11)

The transformation (3.2.3.9-11) is illustrated at Fig. Mapping transformed points. The area of the ellipse, calculated in the new coordinate system is therefore:

$\int_{0}^{1} d γ \int_{0}^{2 π} μ (β, γ) d β = a b \int_{0}^{1} γ d γ \int_{0}^{2 π} d β = \frac{2 π a b}{2} = π a b$	(3.2.3.12)

as expected. The volume of the ellipsoid

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1$	(3.2.3.13)

was calculated at (3.2.1.26) by integrating in two dimensions over this ellipse, with the function

$z = c \sqrt{1 - \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}}$	(3.2.3.14)

As an additional example we'll repeat the integration, but this time in the new coordinate system, that we just used. The expression (3.2.3.14), with the aid of (3.2.3.9), becomes

$z = c \sqrt{1 - \frac{x^{2}}{a^{2}} - \frac{y^{2}}{c^{2}}} = c \sqrt{1 - γ^{2}}$	(3.2.3.15)

and by applying the magnification factor (3.2.3.11), the volume V of the ellipsoid is

$\begin{array}{l} V = 2 a b c \int_{0}^{1} γ \sqrt{1 - γ^{2}} d γ \int_{0}^{2 π} d β = 4 π a b c \int_{0}^{1} u^{2} d u = \frac{4 π}{3} a b c \\ where 1 - γ^{2} = u^{2} \Rightarrow γ d γ = - u d u \end{array}}$	(3.2.3.16)

Parts 1-3

Solution of question 1.

The required moment of inertia is expressed by

$I_{y} = σ \int \int x^{2} d x d y$
The absolute value of the Jacobian (γab) is known from (3.2.3.11) and by use of the required transformation

$x^{2} = γ^{2} a^{2} \cos^{2} β$ One obtains therefore $I_{y} = σ \int \int γ^{2} a^{2} \cos^{2} (β) γ a b d β d γ$

By substituting x and y from the transformation into the given ellipse, we obtain

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = γ^{2} (\cos^{2} β + \sin^{2} β) = γ^{2} = 1$
As a consequence the limits of integration about γ are (0,1) , and the limits of β are (0,2π) , in order to cover the entire ellipse.
The integral becomes therefore

$I_{y} = σ a^{3} b \int_{0}^{2 π} d β \cos^{2} β \int_{0}^{1} γ^{3} d γ = σ a^{3} b π \frac{1}{4} = \frac{π σ a^{3} b}{4}$
where the integrations over β and γ were done simultaneously, since they are not related.

Parts 4-6

Solution of question 2.

The non-convexity of the ellipse with circular holes complicates the direct integration of the moment of inertia. For this reason we'll subtract the contribution of the perforated parts, which are convex regions. For this reason we are going to use the moment of inertia of a perforated disc by taking first its centre at the origin. From the data and the results of part 3, the moment of inertia about the y axis is

$I = \frac{π σ ρ^{4}}{4}$
where we used an ellipse with equal radii.
In order to obtain the moment of inertia of a disc about the y axis, one have to use the Steiner's theorem by use of its real position. This moment of inertia of one disc becomes

$I = \frac{π σ ρ^{4}}{4} + M d^{2} = \frac{π σ ρ^{4}}{4} + σ π ρ^{2} d^{2} = \frac{σ π}{4} (ρ^{4} + 4 ρ^{2} d^{2})$

The final result of the moment of inertia is obtained by subtracting from the result of question 1 (part 3), twice (two discs) the result of part 5, namely

$I_{y} = \frac{σ π}{4} (a^{3} b - 2 ρ^{4} - 8 ρ^{2} d^{2})$

Parts 7-8

Solution of question 3.

The function z transforms to

$z = h (1 - \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}) = h [1 - γ^{2} (\cos^{2} β + \sin^{2} β)] = h (1 - γ^{2})$

By use of the Jacobian and the limits of integration (parts 1 and 2), the required volume becomes

$V = a b h \int_{0}^{2 π} d β \int_{0}^{1} γ (1 - γ^{2}) d γ = 2 π a b h (\frac{1}{2} - \frac{1}{4}) = \frac{π a b h}{2}$
In agreement with the exercise done with Cartesian coordinates.

Parts 9-10

Solution of question 4.

The physical dimensions of the moment of inertia I are

$[I] = Μ Λ^{2}$
where Μ means mass and Λ means length. The dimensions of the density σ are

$[σ] = Μ Λ^{- 2}$ and $[a] = [b] = [ρ] = [d] = Λ$
yielding the correct dimensions for the results of questions 1 and 2.
The physical dimensions of volume are

$[V] = Λ^{3}$ and $[a] = [b] = [h] = [z] = Λ$
yielding the correct dimensions for the result of question 3.

Score

By parts.
All parts are worth 1 point each.

By questions.
Questions 1 and 2 are worth 3 point each.
Questions 3 and 4 are worth 2 points each.

Math Animated™ Mathematical Introduction for Physics and Engineering by Samuel Dagan (Copyright © 2007-2020)