Problem 12 - University Calculus: Early Transcendentals

Question

The area of an ellipse The area \(\pi a b\) of the ellipse \(x^{2} / a^{2}+y^{2} / b^{2}=1\) can be found by integrating the function \(f(x, y)=1\) over the region bounded by the ellipse in the \(x y-\) plane. Evaluating the integral directly requires a trigonometric substitution. An easier way to evaluate the integral is to use the transformation \(x=a u, y=b v\) and evaluate the transformed integral over the disk \(G: u^{2}+v^{2} \leq 1\) in the \(u v\) -plane. Find the area this way.

Step-by-Step Solution

Answer

The area of the ellipse is \(\pi ab\).

Step 1: Apply the Transformation

To simplify the problem, let us use the transformation of variables given by \( x = au \) and \( y = bv \). This changes the region of integration from the ellipse to a unit disk in the \(uv\)-plane, defined by \( u^2 + v^2 \leq 1 \). This is because the transformation scales the axes to unit length.

Step 2: Compute the Jacobian of the Transformation

To change the variables in an integral, we need to determine the Jacobian determinant of the transformation. From \( x = au \) and \( y = bv \), the Jacobian is the determinant of the matrix formed by the partial derivatives:\[ J = \det \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix} = \det \begin{bmatrix} a & 0 \ 0 & b \end{bmatrix} = ab. \]

Step 3: Set Up the Integral in the Transformed Coordinates

The original integral of \( f(x,y) = 1 \) over the ellipse can now be written as an integral over the unit disk \( G \) with the Jacobian included:\[ \int\int_{G} 1 \cdot ab \, dudv = ab \int\int_{G} dudv. \]

Step 4: Evaluate the Integral Over the Unit Disk

The integral \( \int\int_{G} dudv \) over the unit disk \( G: u^2 + v^2 \leq 1 \) represents the area of the unit circle, which is well-known to be \( \pi \). Thus, the integral evaluates to \( \pi \):\[ ab \int\int_{G} dudv = ab \cdot \pi. \]

Step 5: Finalize the Area Calculation

Multiplying by the Jacobian, the area of the ellipse is confirmed to be \( ab \cdot \pi \). Thus, the area of the ellipse expressed by \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is \( \pi ab \).

Key Concepts

Integration Techniques Made Simple

When finding the area of an ellipse, integrating might sound intimidating, but it can be made simple with the right approach.
An ellipse in standard form is expressed as \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), and to find its area using integration, we consider a function \( f(x, y) = 1 \) over this region.
Instead of tackling the direct integration, which involves complex trigonometric substitution, we opt for the variable transformation method.

• This method simplifies calculations by transforming the ellipse to a more familiar shape, such as a unit disk.
• We can then integrate over the simpler shape, avoiding intricate calculations.

The result is not only accurate but also much easier to achieve, making integration far less daunting.
It showcases the beauty of mathematical techniques that turn seemingly complicated problems into simple ones.

Understanding the Jacobian Determinant

The Jacobian determinant is a crucial part of changing variables in integrals.
It essentially acts as a conversion factor that adjusts the scale of our integration when transforming variables.
In our problem, we replace \( x \) and \( y \) with \( au \) and \( bv \), respectively.

Jacobian Calculation

Calculating the Jacobian involves forming a matrix with partial derivatives of those transformations:

• For \( x = au \), the partial derivatives are \( a \) with respect to \( u \) and 0 with respect to \( v \).
• For \( y = bv \), the partial derivatives are 0 with respect to \( u \) and \( b \) with respect to \( v \).

Thus, the Jacobian matrix is \[\begin{bmatrix}a & 0 \0 & b\end{bmatrix}\]And its determinant \( J \) becomes \( ab \).
The Jacobian tells us how much the area changes under the transformation, ensuring our integration accounts for this new scale.

Variable Transformation in Action

Variable transformation is like changing the lenses through which we see a math problem.
In this case, we transform the ellipse into a unit disk, which significantly simplifies the integration process.

Transforming Coordinates

By setting \( x = au \) and \( y = bv \), we move from the original ellipse to a unit disk represented by \( u^2 + v^2 \leq 1 \).
This is because the transformation scales down the ellipse so each axis stretches back to a unit circle in a new coordinate system.

• The process involves substituting these transformed coordinates into the integral.
• Next, multiplying by the Jacobian determinant \( ab \) adjusts for the scale of this new region.

Finally, the integral over the unit disk simplifies to the known area of a circle, \( \pi \).
This transformation turns a complex elliptical region into a straightforward calculation over a circle, demonstrating the elegance of this powerful mathematical strategy.