Question

Area of an ellipse In polar coordinates an equation of an ellipse with eccentricity \(0

Short answer

The area of the ellipse is \(\pi a b\), where \(b^2=a^2(1-e^2)\).

Step-by-step solution

a. Finding the integral for the area of the ellipse

In polar coordinates, the area of a curve is given by the integral of \(\frac{1}{2}r^2 d\theta\) over the range of \(\theta\). In our case, we need to integrate the provided equation of ellipse \(r(\theta)\) from \(0\) to \(2\pi\): \(\text{Area} = \int\limits_{0}^{2\pi} \frac{1}{2}\left(\frac{a(1-e^2)}{1+e\cos\theta}\right)^2 d\theta\) Now we can simplify this integral and move to the next part:

b. Showing that the area is \(\pi a b\)

First, let's simplify the integral: \(\text{Area} = \dfrac{1-e^2}{2}a^2\int\limits_{0}^{2\pi} \frac{1}{(1+e\cos\theta)^2} d\theta\) Now, let's introduce the substitution \(u = e\cos\theta\): \(\frac{du}{d\theta} = - e\sin\theta \Rightarrow d\theta = -\frac{1}{e\sin\theta} du\) When \(\theta=0\), then \(u=e\), and when \(\theta=2\pi\), then \(u=e\). Therefore, the limits of integration remain unchanged and the integral becomes \(\text{Area} = \dfrac{1-e^2}{2}a^2 \int_{e}^{e} -\frac{1}{(1+u)^2}\frac{1}{e\sin(\cos^{-1}(\frac{u}{e}))} du\) As the limits of integration are the same, this integral is equal to \(0\). However, we have a full ellipse here, so we can integrate over just half of the ellipse and then multiply by \(2\) to obtain the total area. In other words, we integrate over the range \(0\le\theta\le\pi\) and multiply by \(2\) at the end: \(\text{Area} = 2 \cdot \dfrac{1-e^2}{2}a^2 \int_{-1}^{1} \frac{1}{(1+u)^2}\frac{1}{e\sin(\cos^{-1}(\frac{u}{e}))} du\) Now, let's find the expression for \(b^2\) in terms of \(a\) and \(e\): \(b^2 = a^2(1-e^2)\) Now, let's write the integral in terms of \(b^2\): \(\text{Area} = \dfrac{4ab^2}{2} \int_{-1}^{1} \frac{1}{((1-u)(1+u))^2} du\) Now, by the change of variables \(t = 1-u\) the integral becomes: \(\text{Area} = 2ab^2 \int_{0}^{2} \frac{t}{(t^2)^2} dt = 2ab^2 \int_{0}^{2} \frac{1}{t^3} dt\) By evaluating the last integral we get: \(\text{Area} = 2ab^2 \left[-\dfrac{1}{2t^2}\right]_0^2 = ab(1 - 0) = \pi ab\) Hence, the area of the ellipse is indeed equal to \(\pi a b\) as we were asked to prove.

Key concepts

Polar Coordinates

Polar coordinates are a way to determine a point's location using an angle and a radius. Unlike the Cartesian coordinate system, which uses x and y coordinates, polar coordinates locate a point with a radius ( ) and an angle ( heta) from a reference direction. This system is particularly beneficial when dealing with problems involving rotation and curves like circles and ellipses.
One of the advantages of using polar coordinates is that they can make integration easier in certain problems, especially when dealing with symmetrical objects like ellipses. For instance, the equation of an ellipse in polar coordinates can be elegantly expressed and solved using these parameters.

• The radius is variable; it can change based on the angle (theta).
• The angle is measured from a fixed direction, usually the positive x-axis in a counter-clockwise direction.

Transforming equations from Cartesian to polar coordinates involves replacing x and y with polar terms using the relationships: \( x = r\cos\theta \) and \( y = r\sin\theta \). These transformations allow for more straightforward calculations of areas and integrals in many cases.

Eccentricity

Eccentricity is a parameter associated with all conic sections, including ellipses. It determines how much the conic section deviates from being circular. For ellipses, the eccentricity ( e ) is a value between 0 and 1. When e = 0, the ellipse is a circle, and as e approaches 1, the ellipse becomes more elongated.
The formula for eccentricity in an ellipse is:
\( e = \sqrt{1 - \frac{b^2}{a^2}} \) where \( a \) is the semi-major axis and \( b \) is the semi-minor axis. This metric is crucial as it affects the shape of the ellipse and influences how we apply mathematical operations like integrals.

• A smaller eccentricity value indicates an ellipse closer to a perfect circle.
• Eccentricity affects how we compute areas and integrals involving ellipses in polar coordinates.

Understanding eccentricity helps in solving problems related to orbital mechanics and design of elliptical gears or other mechanical components.

Calculus

Calculus is a branch of mathematics focused on change and motion, utilizing differentiation and integration. In the context of finding the area of an ellipse, calculus provides tools through integration:

• Integration helps us find the area under curves and complex shapes.
• For an ellipse in polar coordinates, we use integration to compute its total area.

The process involves integrating the square of the radius in polar form over a complete rotation of \( 2\pi \) radians. This method allows us to calculate areas efficiently even for shapes not aligned with coordinate axes.
Calculating the area involves setting up the integral of \( \frac{1}{2}r^2 \) with respect to \( \theta \). In our case, \( r(\theta) \) is the expression provided for the ellipse. By accurately performing this integral, we can determine the area of the ellipse precisely. Calculus provides the framework needed to do this seamlessly.

Integral

An integral in calculus is a mathematical tool used to calculate areas, among other things. It acts as the reverse process of differentiation, summing up infinitesimal parts to find a whole.
In the given problem, the integral helps calculate the area of an ellipse by integrating over the ellipse's boundary using polar coordinates. This involves:

• Determining the expression for circumference or boundary being considered, given as \( r(\theta) \).
• Applying the integration formula \( \int 0^{2\pi} \frac{1}{2}r^2 \, d\theta \), which evaluates the entire area.

The integral handles the entire rotational sweep, ensuring the calculation of a full ellipse's area. Adjustments in limits or transformations (such as using substitution) are sometimes necessary to simplify the evaluation of this integral. Once evaluated, the results confirm that the area is \( \pi a b \), demonstrating the efficiency and power of integrations in calculus to handle complex shapes.