Question

Let \(R\) be the region bounded by the ellipse \(x^{2} / a^{2}+y^{2} / b^{2}=1,\) where \(a > 0\) and \(b>0\) are real numbers. Let \(T\) be the transformation \(x=a u, y=b v\). Find the center of mass of the upper half of \(R(y \geq 0)\) assuming it has a constant density.

Short answer

Answer: The center of mass coordinates for the upper half of the region R are \((\bar{x}, \bar{y}) = (0, \frac{2b}{\pi})\).

Step-by-step solution

Find the area of the upper half of the ellipse

First, we need to find the area of the upper half of the ellipse. For that, we can use the transformation \(T\) and calculate the area in terms of \(u\) and \(v\). As \(x=au\) and \(y=bv\), we can find the Jacobian of the transformation. The Jacobian matrix is given as: $$ J=\begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \\ \end{bmatrix} = \begin{bmatrix} a & 0 \\ 0 & b \\ \end{bmatrix} $$ The determinant of the Jacobian matrix is \(|J|=ab\). Now, to find the area of the upper half of the ellipse, we transform the ellipse equation in terms of \(u\) and \(v\): $$ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \implies u^2 + v^2 =1 $$ So, the area \(A\) of the upper half of the ellipse is given by the following integral in \(u, v\) plane with respect to r, where \(r\) denotes the upper half of the unit circle: $$ A = \int_{r} |J| \, du\,dv = \int_{r} ab \, du\,dv $$ As the upper half of the unit circle can be described by the parameterization \(u=\cos(\theta)\), \(v=\sin(\theta)\), with \(0 \leq \theta \leq \pi\), the area integral becomes: $$ A = ab \int_{0}^{\pi} d\theta = \pi ab $$

Calculate \(\bar{x}\) and \(\bar{y}\)

Now we can calculate \(\bar{x}\) and \(\bar{y}\). For \(\bar{x}\), we have: $$ \bar{x} = \frac{1}{A} \int_{r} x |J| \, du\,dv = \frac{1}{\pi ab} \int_{r} (au) ab \, du\,dv = \frac{a^{2}b}{\pi ab} \int_{r} u \, du\,dv $$ Now we can change the variables to polar coordinates as in step 1 using the parameterization \(u=\cos(\theta)\), \(v=\sin(\theta)\), so we get: $$ \bar{x} = \frac{a}{\pi} \int_{0}^{\pi} \cos(\theta) d\theta = \frac{a}{\pi}[ \sin(\theta) ]_{0}^{\pi} = 0 $$ For \(\bar{y}\), we get: $$ \bar{y} = \frac{1}{A} \int_{r} y |J| \, du\,dv = \frac{1}{\pi ab} \int_{r} (bv) ab \, du\,dv = \frac{ab^{2}}{\pi ab} \int_{r} v \, du\,dv $$ Using the same parameterization as before, we get: $$ \bar{y} = \frac{b}{\pi} \int_{0}^{\pi} \sin(\theta) d\theta = \frac{b}{\pi}[-\cos(\theta)]_{0}^{\pi} = \frac{2b}{\pi} $$

Present the center of mass

The center of mass coordinates \((\bar{x}, \bar{y})\) of the upper half of the ellipse \(R(y \geq 0)\) are: $$ (\bar{x}, \bar{y}) = (0, \frac{2b}{\pi}) $$

Key concepts

Ellipses

Ellipses are fascinating shapes that have many applications in mathematics and physics. They are defined as the set of all points for which the sum of the distances to two fixed points, called foci, is constant. The equation of an ellipse in its standard form is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), where \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes respectively. When \(a > b\), the ellipse is longer horizontally, and when \(b > a\), it is longer vertically.
Ellipses are symmetric about their axes, and this symmetry makes them an interesting subject of study in center of mass problems. For any point on the ellipse, a transformation can sometimes simplify complex calculations by changing the coordinate system. Understanding ellipses deeply involves knowing their properties like eccentricity, parametric equations, and how to apply transformations to them.

Jacobians

The Jacobian is an essential tool in changing variables for multi-variable integrals. It helps in transforming integrals from one coordinate system to another, making calculations easier. The Jacobian is defined as the determinant of the Jacobian matrix, which consists of partial derivatives of the transformation functions.
In our context, the Jacobian matrix for the transformation \(x=au, y=bv\) is given by:

• \(\frac{\partial x}{\partial u} = a\)
• \(\frac{\partial x}{\partial v} = 0\)
• \(\frac{\partial y}{\partial u} = 0\)
• \(\frac{\partial y}{\partial v} = b\)

So, the Jacobian matrix is:
\[ J = \begin{bmatrix} a & 0 \ 0 & b \ \end{bmatrix} \]
The determinant or the Jacobian is then \(|J| = ab\). This value \(|J|\) is crucial when computing the new bounds and integrals in the transformed coordinates. It essentially scales the differential area element when the grid of coordinates is deformed in the transformation.

Density

Density plays a fundamental role in determining the center of mass of an object. Mathematically, density is defined as mass per unit area (for two-dimensional objects) or mass per unit volume (for three-dimensional objects). Here, we assume a constant density, which simplifies the calculations because we do not need to integrate over a variable density function.
When density is constant over an area, computations related to the center of mass become simpler, as they are directly proportional to the geometric properties of the region, rather than the varying mass distribution. This allows us to focus on the shape and boundaries of the region, which in our problem is half of an ellipse.
In more complex scenarios where the density varies, the center of mass calculations would require integrating the density function over the given shape, adding complexity to the integration and potentially necessitating numerical methods for solutions.

Coordinate Transformation

Coordinate transformation is a useful technique in mathematics that allows us to convert complex regions into simpler, often standard geometric shapes for easier analysis. In this exercise, we employ the transformation \(x=au\) and \(y=bv\), effectively normalizing the ellipse to a unit circle \(u^2 + v^2 = 1\).
This conversion is done because working with a unit circle is generally more straightforward due to its symmetry and simpler bounds. The transformation aligns the axes of the ellipse with the standard coordinate axes, making integrals over these regions less complicated.

• Reduces the complexity of the shape.
• Facilitates the application of polar coordinates.
• Makes it easier to apply analytic methods.

Coordinate transformations are heavily used in fields like physics and engineering, where they allow us to work with a problem in a coordinate system best suited to the problem's symmetry and boundaries, enhancing both comprehension and solution simplicity.