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: \((0, \frac{2b}{3\pi})\).

Step-by-step solution

Determine the area of the upper half of the ellipse

The area of the full ellipse is given by \(A_{full}=\pi ab\). Since we want only the area of the upper half of the ellipse, we divide the full area by 2. So, the area of the upper half is: \(A = \frac{1}{2}(\pi ab)\).

Prepare to transform the integral from \((x, y)\) to \((u, v)\) coordinates using the Jacobian

The Jacobian of the transformation \(x=au\) and \(y=bv\) is given by \(\det \left(\frac{\partial(x, y)}{\partial(u, v)}\right)\). Calculate the Jacobian as follows (using the partial derivatives of \(x\) and \(y\) with respect to \(u\) and \(v\)): $J = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} a & 0 \\ 0 & b \end{vmatrix} = ab$.

Calculate the integrals for \(\bar{x}\) and \(\bar{y}\) in the \((u, v)\) coordinate system

For the upper half of ellipse \(R\) in the \((u, v)\) coordinate system, we will have \(u^2 + v^2 \leq 1\) and \(v \geq 0\). We can rewrite the center of mass coordinates as: \(\bar{x} = \frac{1}{A} \iint_R x \, dA = \frac{1}{\pi ab}\iint_R (au) \, dA(u, v)\) \(\bar{y} = \frac{1}{A} \iint_R y \, dA = \frac{1}{\pi ab}\iint_R (bv) \, dA(u, v)\) Using our transformation and Jacobian, these become: \(\bar{x} = \frac{1}{\pi}\int_0^1 u \left( \int_0^{\sqrt{1-u^2}} (ab) \, dv \right)\, du\) \(\bar{y} = \frac{1}{\pi}\int_0^1 v \left( \int_{-v}^v (ab) \, du \right)\, dv\)

Evaluate the integrals to find \(\bar{x}\) and \(\bar{y}\)

Evaluate the integrals, and we obtain the following results: \(\bar{x} = \frac{1}{\pi}\int_0^1 u \left( ab \cdot 2\sqrt{1-u^2} \right) \, du = \frac{2ab}{\pi}\int_0^1 u\sqrt{1-u^2} \, du = 0\) \(\bar{y} = \frac{1}{\pi}\int_0^1 v \left( ab \cdot 2v \right) \, dv = \frac{2ab}{\pi}\int_0^1 v^2 \, dv = \frac{2ab}{3\pi}\)

Find the center of mass in the original \((x, y)\) coordinate system

The center of mass in the \((u, v)\) coordinate system is \((\bar{u}, \bar{v}) = (0, \frac{2}{3\pi})\). Transforming back into the \((x, y)\) coordinate system, we obtain: \(\bar{x} = a\bar{u} = 0\) \(\bar{y} = b\bar{v} = \frac{2b}{3\pi}\) The center of mass of the upper half of the region \(R\) is \(\left(\bar{x}, \bar{y}\right) = \boxed{\left(0, \frac{2b}{3\pi}\right)}\).

Key concepts

Double Integrals

Double integrals play a crucial role in calculus, particularly when we want to calculate the area, volume, or mass of a region in two dimensions. They extend the concept of a single integral, allowing us to integrate over a two-dimensional area. When finding the center of mass for a region like an ellipse, we use double integrals to integrate over its entire area. In simple terms, we first integrate with respect to one variable while keeping the other fixed, and then integrate the result with respect to the second variable.

The use of double integrals in finding the center of mass involves integrating the product of the density function and the coordinates, which gives us a measure of the 'weighted average' position of mass across the region. The results of these integrals provide the coordinates of the center of mass, \(\bar{x}\) and \(\bar{y}\), which indicate the balance point of the region assuming a uniform density. The calculation can become complex, but when dealing with symmetric regions, some integrals simplify due to symmetry, as seen in the solution for \(\bar{x}\) which evaluates to zero.

Elliptical Region

An elliptical region is a set of points on a plane that form the shape of an ellipse. An ellipse can be thought of as a stretched circle, described by the general equation \(x^2/a^2 + y^2/b^2 = 1\) where \(a\) and \(b\) represent the semi-major and semi-minor axes, respectively. When solving problems involving elliptical regions, properties such as symmetry can greatly simplify calculations.

In the context of finding the center of mass, the elliptical region’s symmetry around its axes comes in handy. For example, the center of mass along the x-axis (\bar{x}) for the upper half of an ellipse lies on the y-axis due to this symmetry. This natural balance means that, regardless of how dense the material is, as long as it's uniformly distributed, \(\bar{x}\) will always be at the center of the ellipse along the x-axis, which simplifies to zero if the ellipse is symmetric along the y-axis.

Jacobian Transformation

When dealing with double integrals, especially over regions like an ellipse, direct integration can be challenging. Here, the Jacobian transformation comes to the rescue. The Jacobian transformation is a mathematical tool that allows us to transform a difficult set of integration variables into a more manageable one by changing the coordinate system.

The Jacobian itself, denoted by \(J\), is the determinant of the matrix of partial derivatives and represents how much the area element \(dA\) is stretched or shrunk during the transformation. In our exercise, by going from the coordinates (\(x, y\)) to (\(u, v\)) such that \(x=au\) and \(y=bv\), we simplify the region of integration to one bounded by a unit circle, which is far easier to handle. The resultant Jacobian is \(ab\), which compensates for the area distortion during transformation, ensuring the integral over the new region gives the correct values for the original problem.

Center of Mass Coordinates

The center of mass of an object is the point at which the mass distribution of the object is balanced. Its coordinates are found by calculating the weighted average positions of all the mass elements distributed across the object. For a continuous mass distribution over a region in a plane with uniform density, the coordinates \(\bar{x}\) and \(\bar{y}\) of the center of mass are found using double integrals that account for the entire mass and its distribution.

In our example, we assumed a constant density, which simplified the calculation by removing density from the integrals. The center of mass coordinates are linked directly to the geometric properties of the region: \(\bar{x}\) collapses to zero due to symmetry, and \(\bar{y}\) is found by integrating over the positive half of the region. The pivotal step of transforming the coordinate system to one that makes the region easier to integrate over is facilitated by the Jacobian, and the resulting integrals reveal the center of mass coordinates for the given elliptical region.