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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 area of \(R.\)

Short Answer

Expert verified
Answer: The area of the region R enclosed by the ellipse is \(A(R) = \pi ab\).

Step by step solution

01

Identify the equation and the transformation given

We have the equation of the ellipse \(x^{2} / a^{2}+y^{2} / b^{2}=1,\) and the transformation \(x=au, y=bv\).
02

Express equation in terms of transformation variables

Replace x with \(au\) and y with \(bv\) in the ellipse equation: \((au)^{2} / a^{2}+(bv)^{2} / b^{2}=1\). This simplifies to \(u^2+v^2=1\) which is the equation of a unit circle.
03

Calculate the Jacobian of the transformation

The Jacobian determinant of the transformation (\(x = au, y = bv\)) is given by: $\frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} a & 0 \\ 0 & b \end{vmatrix} = ab$ Since \(a>0\) and \(b>0\), the Jacobian determinant is positive and nonzero.
04

Choose limits of integration and integrate

Since we need to integrate over the unit circle, the limits of integration for the variables u and v will be the entire circular region, enclosed by \(u^2 + v^2 = 1\). We will use polar coordinates for this: change \((u,v)\) to \((r\cos\theta, r\sin\theta)\) and \(0 \leq r \leq 1\), \(0 \leq \theta \leq 2\pi\). Now, we express the Jacobian determinant in polar coordinates: $\frac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} a\cos\theta & -ar\sin\theta \\ b\sin\theta & br\cos\theta \end{vmatrix} = abr$ The area of the ellipse can now be calculated as the double integral of the Jacobian times the differential area element \(dA\): \(A(R)=\int_{0}^{2\pi} \int_{0}^{1} abr \, dr \, d\theta\)
05

Evaluate the integral

First, integrate with respect to r: \(\int_{0}^{1} abr \, dr = \frac{1}{2}abr^2\Bigr|_0^1 = \frac{1}{2}ab\) Now, integrate with respect to \(\theta\): \(\int_{0}^{2\pi} \frac{1}{2}ab \, d\theta = \frac{1}{2}ab\theta\Bigr|_0^{2\pi} = \pi ab\) So, the area of the region \(R\) enclosed by the ellipse is \(A(R) = \pi ab\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Transformation of Axes
The term "Transformation of Axes" is used to describe how we manipulate the coordinate system to simplify a problem, often leading to easier calculations or a clearer understanding of the situation. In the context of the ellipse provided in the exercise, we conduct an axis transformation using the equations: \(x = au\) and \(y = bv\). By substituting these into the ellipse equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we turn what was initially an elliptical equation into the unit circle equation \(u^2 + v^2 = 1\).
This transformation simplifies our integration process because working with circles is generally easier than with ellipses.
### Benefits of Axis Transformation:
  • Simplifies the geometry of the problem.
  • Transforms complex shapes into simpler ones.
  • Facilitates the use of integration in different coordinate systems.
Transformations like this are fundamental in calculus because they allow us to exploit symmetry and geometric simplicity to perform calculations that would otherwise be cumbersome.
Jacobian Determinant
The Jacobian Determinant is a mathematical expression used to understand how much a transformation scales, moves, or skews a given region. When we transform variables, as we did with \(x = au\) and \(y = bv\), we need to find the Jacobian to properly adjust for the change of variables.

In our specific case, the Jacobian determinant describes the area scaling effect when moving from the \((x, y)\) space to the \((u, v)\) space. We calculate it using partial derivatives:
\[\frac{\partial(x, y)}{\partial(u, v)} = \begin{vmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{vmatrix} = \begin{vmatrix}a & 0 \0 & b\end{vmatrix} = ab\]### Key Points of Jacobian Determinants:
  • They are crucial for changing variables in integration.
  • They extend to multiple dimensions, not just from 2D spaces.
  • The determinant gives the scaling factor for area (or volume in higher dimensions).
Learning to compute the Jacobian correctly ensures that we account for the changes in "size" that occur with transformations, ensuring the integral remains valid.
Integration in Polar Coordinates
Polar Coordinates are an alternative to the more common Cartesian Coordinates (x, y).
In cases involving symmetry or circular shapes, like our transformed ellipse equation, polar coordinates simplify integration. In our exercise, after transforming the ellipse equation into the unit circle \(u^2 + v^2 = 1\), we opt to integrate in polar coordinates. This involved setting \(u = r\cos\theta\) and \(v = r\sin\theta\), which translates our circle into polar equations.
The limits for \(r\) are from 0 to 1, and \(\theta\) spans from 0 to \(2\pi\). The differential area element, represented in polar coordinates, is \(r \, dr \, d\theta\).### Why Use Polar Integration:
  • Circular or radial symmetry simplifies to straight lines in polar coordinates.
  • Integration is made easier for circular paths or areas.
  • The polar coordinate form adapts well to the integration process using radial distance and angle.
This makes polar coordinates an effective method for integration, especially where circular shapes or radial symmetry are present in the problem. Using polar integration made it possible to compute the area of our ellipse efficiently, leading to the solution \(\pi ab\), highlighting the power and utility of polar transformation.

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