Question

The upper half of the ellipse centered at the origin with axes of length \(2 a\) and \(2 b\) is described by \(y=\frac{b}{a} \sqrt{a^{2}-x^{2}}\) (see figure). Find the area of the ellipse in terms of \(a\) and \(b\)

Short answer

Answer: The area of an ellipse with major axis "a" and minor axis "b" is abπ.

Step-by-step solution

Set up the integral for the upper half of the ellipse

The area of the upper half of the ellipse can be found using the integral of the equation: $$ A_{upper} = \int_{-a}^{a} \frac{b}{a}\sqrt{a^2-x^2}\, dx $$

Use substitution method

To make this integral easier to solve, use substitution method. Let \(x = a\sin{u}\), so \(dx = a\cos{u}\, du\). Now, we need to change the integration limits. When \(x = -a\), then \(u = -\frac{\pi}{2}\) and when \(x = a\), then \(u = \frac{\pi}{2}\). So, our new integral becomes: $$ A_{upper} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{b}{a}\sqrt{a^2-(a\sin{u})^2}\, (a\cos{u}\, du) $$

Simplify the integral

The integral can now be simplified: $$ A_{upper} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} b\sqrt{a^2(1-\sin^2{u})}\,a\cos{u}\, du = ab\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{1-\sin^2{u}}\,a\cos{u}\, du $$ Since \(\sqrt{1-\sin^2{u}} = \cos{u}\), the integral becomes: $$ A_{upper} = ab\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2{u}\, du $$

Use the double-angle identity

The double-angle identity for cosine is \(\cos^2{u}=\frac{1+\cos{2u}}{2}\). Use this to rewrite the integral: $$ A_{upper} = ab \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\cos{2u}}{2}\, du $$

Integrate

Now, integrate the function with respect to \(u\): $$ A_{upper} = ab \left[ \frac{1}{2}u + \frac{1}{4}\sin{2u}\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} $$ Evaluate the integral at the upper and lower limits: $$ A_{upper} = ab\left[ \left( \frac{\pi}{4} + 0\right) - \left(-\frac{\pi}{4} + 0\right) \right] = ab \cdot \frac{\pi}{2} $$

Find the area of the entire ellipse

Now that we have the area of the upper half of the ellipse, we can find the area of the entire ellipse by multiplying this result by 2: $$ \text{Area of ellipse} = 2A_{upper} = 2ab \cdot \frac{\pi}{2} = ab\pi $$ The area of the ellipse is \(ab\pi\).

Key concepts

Integration Techniques

When calculating the area under a curve, integration is a crucial method. The elliptical area problem uses this technique to integrate the function that represents the upper half of the ellipse. By setting up the integral

• From \(-a\) to \(a\), we accumulate the small areas underneath the curve \(y=\frac{b}{a}\sqrt{a^{2}-x^{2}}\).

Integrating this expression helps us find the area of half of the ellipse. This is a classic application of integration in calculus to derive a physical quantity. Understanding how to set up and solve these kinds of integrals is vital for different branches of science and engineering.
The complex part starts when you encounter square roots or powers like in our function. Integration techniques including substitution (which is used in this example) simplify otherwise complicated integrals.

Substitution Method

In this problem, the substitution method simplifies the integration process. It's a magical tool in calculus that transforms complex integrals into simpler forms. Here, let \(x = a\sin{u}\) which changes the integration from variable \(x\) to \(u\).

• When \(x = \pm a\), \(u\) varies from\(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).
• The \(dx=a\cos{u}\,du\), expressing the differential equation in terms of \(u\), is crucial for implementing substitution.

This transformation is particularly useful because it often transforms complicated integrals into more familiar forms, or as in this case, makes it possible to use trigonometric identities to further simplify the calculation. Substitution effectively converts the ellipse integration into a simpler problem, making the process manageable.

Trigonometric Identities

Trigonometric identities are handy tools for solving integrals involving trigonometric functions. In finding the ellipse's area, they simplify the expression obtained from the substitution method. A key identity here is:

• \( \cos^{2}{u} = \frac{1+\cos{2u}}{2} \)

This identity is critical because it allows the \(\cos^2{u}\) term to transform into a more integrable form. By using the double-angle identity, you break down and simplify another layer of complexity in the integral.
This technique helps in evaluating \(\int \cos^2{u}\, du\) emerging from the integration after substitution. Successfully tackling the integral with identities like this is a key part of problem-solving in calculus, reinforcing their utility in simplifying expressions.

Calculus

Calculus provides the framework for finding and defining changes systematically. In the problem of calculating the ellipse area, calculus helps produce precise measurements using integration.

• A significant step is understanding how calculus translates geometric figures into mathematical expressions and vice versa.
• It bridges initial algebraic problems with geometric solutions, as seen when solving for the ellipse’s area.

By mastering calculus, you gain the ability to work through complex mathematical problems efficiently. In fact, the entire procedure of converting the problem, utilizing the substitution and identifying applicable trigonometric identities stems from the principles of calculus.
Through calculus, we derive meaningful results that are applicable in various fields related to physics, engineering, and beyond. It emphasizes the importance of creating a solid foundation in calculus to effectively calculate, analyze, and comprehend complex systems and phenomena.