Problem 317 - Calculus Volume 1

Question

The area of the top half of an ellipse with a major axis that is the \(x\) -axis from \(x=-1\) to \(a\) and with a minor axis that is the \(y\) -axis from \(y=-b\) to \(b\) can be written as \(\int_{-a}^{a} b \sqrt{1-\frac{x^{2}}{a^{2}}} d x .\) Use the substitution \(x=a \cos t\) to express this area in terms of an integral of a trigonometric function. You do not need to compute the integral.

Step-by-Step Solution

Answer

Integrate \( ab \sin^2 t \) from 0 to \( \pi \).

Step 1: Understanding the Problem

We have an ellipse with major and minor axes along the x and y-axes, respectively. The goal is to express the area of the top half, from \( x = -a \) to \( x = a \), using trigonometric functions via substitution.

Step 2: Setup the Substitution

Using the given substitution \( x = a \cos t \), differentiate both sides to find \( dx \). So the substitution \( x = a \cos t \) gives \( dx = -a \sin t \, dt \).

Step 3: Changing Limits of Integration

The original limits for \( x \) are from \( -a \) to \( a \). For \( x = -a \), \(\cos t = -1\) implying \( t = \pi \). For \( x = a \), \( \cos t = 1 \) implying \( t = 0 \). The new limits are \( t = \pi \) to \( t = 0 \).

Step 4: Substitute into the Integral

Substitute \( x = a \cos t \) and \( dx = -a \sin t \, dt \) into the integral, resulting in: \[ \int_{\pi}^{0} b \sqrt{1-\frac{(a \cos t)^2}{a^2}} \cdot (-a \sin t) \, dt \].

Step 5: Simplify the Integral

Simplify inside the square root: \[ \sqrt{1 - \cos^2 t} = \sin t \]. Thus, the integral becomes: \[ \int_{\pi}^{0} b (-a \sin^2 t) \, dt \].

Step 6: Simplify and Adjust Limits

Change the limits from \( t = \pi \) to \( 0 \) by adjusting signs, resulting in: \[ \int_{0}^{\pi} ab \sin^2 t \, dt \].

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a valuable technique in solving integrals, especially those involving square roots of quadratic expressions. It simplifies complex expressions by exploiting relationships in trigonometry. In our exercise, the substitution \( x = a \cos t \) smoothly transforms the integrand into a simpler form. This substitution harnesses the identity \( \cos^2 t + \sin^2 t = 1 \) to simplify the square root within the integral. The substitution effectively reduces the complexity of the problem.
This process requires recalculating the differential \( dx \), which becomes \( -a \sin t \, dt \). This new form allows easier integration along a different variable and ultimately leads to a more streamlined solution method. Substitutions like \( x = a \cos t \) are common when dealing with circles and ellipses due to their reliance on trigonometric identities.

Integral Calculus

Integral calculus is the branch of mathematics that deals with accumulation of quantities, such as areas under curves. In the context of ellipses, we often aim to find the area bound by its perimeter. Here, calculus plays a crucial role. The integral \( \int_{-a}^{a} b \sqrt{1-\frac{x^{2}}{a^{2}}} \, dx \) represents the area calculation of the half-ellipse over the \( x \)-axis.
The integral symbolizes the sum of all infinitesimal slices of the ellipse's top half. By transforming this expression using trigonometric substitution, we make it more digestible. Integral calculus provides the tools to deal with continuous change and accumulation, turning complex shapes like ellipses into manageable curves for computation.

Limits of Integration

Limits of integration define the interval over which we compute the integral. They are crucial in determining the specific portion of a function we are analyzing. In the original problem, the limits \( x = -a \) to \( x = a \) specify the section of the ellipse along the x-axis. After substitution with \( x = a \cos t \), these limits translate to trigonometric limits \( t = \pi \) to \( t = 0 \).
The transformation of limits results from solving \( \cos t \) to match the boundary values of \( x \). It reflects how the function's domain shifts when viewed through a different variable. Adjusting these limits ensures the integral represents the same geometry as initially intended.

Polar Coordinates

Polar coordinates offer a method to describe curves using distances and angles instead of Cartesian \((x, y)\) coordinates. They simplify trigonometric integrations and are particularly apt for circular and elliptical geometries. While the original problem primarily involves trigonometric substitution, the move towards expressions like \( x = a \cos t \) aligns with polar concepts where a point's position depends on an angle and a distance.
Ellipses and related conic sections often benefit from polar or parametric interpretations. This aligns with the way ellipses distribute around a focus, frequently analyzed using angles and radii. These approaches not only simplify calculus problems involving circles and ellipses but also enhance our understanding by aligning mathematical solutions with geometric intuition.