Problem 11 - Thomas Calculus

Question

Polar moment of inertia of an elliptical plate \(A\) thin plate of constant density covers the region bounded by the ellipse \(x^{2} / a^{2}+y^{2} / b^{2}=1, a>0, b>0,\) in the \(x y\) -plane. Find the first moment of the plate about the origin. (Hint: Use the transformation \(x=a r \cos \theta, y=b r \sin \theta . )\)

Step-by-Step Solution

Answer

The first moment about the origin is \(\frac{2\pi}{3} \rho ab\).

Step 1: Understanding the Transformation

First, we recognize that the plate is bounded by the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) in the Cartesian coordinates. The proposed transformation \(x = a r \cos \theta\) and \(y = b r \sin \theta\) converts this into polar coordinates where \(0 \leq r \leq 1\) and \(0 \leq \theta < 2\pi\). This transformation will simplify integration over the region.

Step 2: Setting Up the Jacobian

We need to find the Jacobian of the transformation to express \(dx \times dy\) in terms of \(dr \times d\theta\). The transformation gives \(x = a r \cos \theta\) and \(y = b r \sin \theta\), resulting in the Jacobian determinant \( J = \left| \frac{\partial (x, y)}{\partial (r, \theta)} \right| = ab r\). This tells us that \(dx \, dy = ab r \, dr \, d\theta\).

Step 3: Calculate the First Moment about the Origin

To find the first moment of the plate about the origin, we integrate the product of the density function \(\rho\) and the radius \(r\), expressed by \(M_0 = \int_0^{2\pi} \int_0^1 \rho \cdot ab r^2 \, dr \, d\theta\). Assuming constant density \(\rho\) simplifies to \(\rho ab \int_0^{2\pi} \int_0^1 r^2 \, dr \, d\theta\).

Step 4: Perform Double Integration

Integrate with respect to \(r\) and then \(\theta\). The integration \(\int_0^1 r^2 \, dr = \left. \frac{r^3}{3} \right|_0^1 = \frac{1}{3}\). Thus, the inner integral becomes \(\rho ab \times \frac{1}{3}\). Continuing with \(\int_0^{2\pi} 1 \, d\theta = 2 \pi\), gives the total integral of \(\rho ab \times \frac{1}{3} \times 2\pi = \frac{2\pi}{3} \rho ab\).

Step 5: Summarizing the Result

The first moment of the elliptical plate about the origin, considering the constant density \(\rho\), is \(\frac{2\pi}{3} \rho ab\). This result reflects the symmetric nature and uniform density distributed over the elliptical region.

Key Concepts

Ellipse

An ellipse is a geometric shape that looks like a flattened circle. It's defined by its major and minor axes, which are lines that run through the longest and shortest parts of the ellipse, respectively. In mathematical terms, an ellipse can be represented by the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \( a \) and \( b \) are constants that determine the lengths of the semi-major and semi-minor axes. Here, \( a > 0 \) and \( b > 0 \), ensuring the shape is an ellipse.

Understanding the shape of the ellipse is crucial because it helps us apply transformations later on in the problem. This transformation simplifies how we calculate areas and moments.

Polar Coordinates

Polar coordinates are a way of mapping points in a plane using a radius and an angle, as opposed to traditional Cartesian coordinates which use an \( x \)-value and a \( y \)-value. In polar coordinates, a point is described as \( (r, \theta) \), where \( r \) is the distance from the origin to the point, and \( \theta \) is the angle from the positive \( x \)-axis.

In this problem, converting the ellipse equation into polar coordinates makes integration easier. The transformation equations are \( x = a r \cos \theta \) and \( y = b r \sin \theta \). As a result, the boundary \( 0 \leq r \leq 1 \) and \( 0 \leq \theta < 2\pi \) represent the ellipse in polar form.

Jacobian

The Jacobian is a mathematical concept used in changing variables of multiple integrals. It provides the scale factor by which we should adjust our differential area element as the variables change.

In this exercise, when transitioning from Cartesian to polar coordinates, we need to find the Jacobian determinant for the transformation \( (x, y) \to (r, \theta) \). The expressions \( x = a r \cos \theta \) and \( y = b r \sin \theta \) give us the Jacobian \( J = \left| \frac{\partial (x, y)}{\partial (r, \theta)} \right| = ab r \). Consequently, the area element \( dx \, dy \) becomes \( ab r \, dr \, d\theta \) after applying the Jacobian.

Integration

Integration is a fundamental concept in calculus that helps compute areas, volumes, and other quantities under curves. It’s like summing up infinite small parts to find a whole.

In this problem, we perform a double integration to find the first moment of the plate about the origin using the polar form of the ellipse. The integration problem is simplified by transforming coordinates, allowing straightforward integration over \( r \) and \( \theta \). The double integral is set up as follows:

• First, integrate with respect to \( r \) over \( [0,1] \), giving \( \int_0^1 r^2 \, dr = \frac{1}{3} \).
• Then integrate with respect to \( \theta \) over \( [0, 2\pi] \), yielding \( \int_0^{2\pi} 1 \, d\theta = 2\pi \).

Combining these results, the total integral is \( \frac{2\pi}{3} \rho ab \).

Constant Density

Density is the amount of mass per unit volume. When an object has constant density, it means that its mass is evenly distributed across its entire surface.

In this exercise, the elliptical plate is described as having a constant density \( \rho \). This simplifies the calculation because we can treat the density as a constant factor outside of the integration. When we calculate the first moment of the plate, we consider the product of the density and the calculated area. Hence, the first moment is based on \( \rho ab \int_0^{2\pi} \int_0^1 r^2 \, dr \, d\theta \) which gives the result \( \frac{2\pi}{3} \rho ab \), reflecting the uniformity of the elliptical plate's density distribution.