Problem 44 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate \(\iint_{S} x^{2} d A\), where \(S\) is the region between the ellipse \(x^{2}+2 y^{2}=4\) and the circle \(x^{2}+y^{2}=4\)

Step-by-Step Solution

Answer

The answer is the integral evaluates to a numerical result after the last step.

Step 1: Identify region S

The region \( S \) lies between the ellipse \( x^2 + 2y^2 = 4 \) and the circle \( x^2 + y^2 = 4 \). To analyze this, note that both curves are enclosed regions with the same boundary value for their respective equations.

Step 2: Convert to polar coordinates

Since both integral curves are centered at the origin, it is useful to convert Cartesian coordinates to polar coordinates. Recall the conversions: \( x = r \cos \theta \), and \( y = r \sin \theta \). The limits for \( r \) will depend on \( \theta \): from the ellipse, \( r = \frac{2}{\sqrt{\cos^2 \theta + 2 \sin^2 \theta}} \) and for the circle, \( r = 2 \).

Step 3: Set up the polar integral

With these conversions, the double integral can be written as \( \int_0^{2\pi} \int_{\frac{2}{\sqrt{\cos^2\theta + 2\sin^2\theta}}}^{2} r^2 \cos^2 \theta \cdot r \, dr \, d\theta \). Notice how the area element \( dA = r \, dr \, d\theta \) and function \( x^2 = r^2 \cos^2 \theta \) are rewritten in polar form.

Step 4: Evaluate the integral

The integral becomes \( \int_0^{2\pi} \int_{\frac{2}{\sqrt{\cos^2\theta + 2\sin^2\theta}}}^{2} r^3 \cos^2 \theta \, dr \, d\theta \). Separate the internal integral by \( \theta \). After finding the indefinite integral for \( r \), we evaluate it from the ellipse's limit to the circle's limit.

Step 5: Perform radial integration

Compute the integral with respect to \( r \): \( \frac{r^4}{4} \cos^2 \theta \). Evaluate from the ellipse radius to the circle radius: \( \left[ \frac{16}{4} \cos^2 \theta - \frac{\left(\frac{2}{\sqrt{\cos^2 \theta + 2 \sin^2 \theta}}\right)^4}{4} \cos^2 \theta \right] \).

Step 6: Substitute and simplify

Substitute back the expression for the ellipse's radius, \( \frac{2}{\sqrt{\cos^2 \theta + 2 \sin^2 \theta}} \). This step results in the expression: \( \int_0^{2\pi} \left[ 4 \cos^2 \theta - 1 \right] \cos^2 \theta \, d\theta \). Simplify by treating the trigonometric identity and properties.

Step 7: Final integration

Integrate the simplified expression over \( \theta \): \( \int_0^{2\pi} \left( 4 \cos^4 \theta - \cos^2 \theta \right) \, d\theta \). This can be approached using trigonometric identities (e.g., \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \) and \( \cos^4 \theta = \left( \frac{1 + \cos 2\theta}{2} \right)^2 \)) to simplify, then integrate.

Key Concepts

Polar Coordinates

Polar coordinates are a two-dimensional coordinate system where each point on a plane is defined by a distance from a reference point and an angle from a reference direction. Unlike Cartesian coordinates, which use coordinates \(x\) and \(y\), polar coordinates use coordinates \(r\) (the radius) and \(\theta\) (the angle). This system is especially useful in problems involving symmetry around a point, such as circles and ellipses, as it can simplify calculations.

• The conversion formulas from Cartesian to polar are \(x = r \cos \theta\) and \(y = r \sin \theta\), where \(r = \sqrt{x^2+y^2}\) and \( heta = \arctan(\frac{y}{x})\).
• When switching to polar coordinates in integration, the area element \(dA\) in Cartesian becomes \(r \, dr \, d\theta\). This accounts for the variable areas swept out by different values of \(r\) and \(\theta\).

In this exercise, converting from Cartesian to polar coordinates facilitates the evaluation of the integral because the region of integration is centered around the origin and involves circular and elliptical shapes, making polar coordinates a natural fit.

Ellipse and Circle Regions

The problem involves the region between an ellipse and a circle, both centered at the origin. Understanding these regions geometrically can simplify integration tasks.

• The circle given by \(x^2 + y^2 = 4\) is a perfect circle with a radius of 2, centered at the origin. This equation is symmetric in both axes and hence, in any direction.
• The ellipse given by \(x^2 + 2y^2 = 4\) is also centered at the origin but is stretched along the \(y\)-axis due to the factor of 2. To see this clearly, rewrite it as \(\frac{x^2}{4} + \frac{y^2}{2} = 1\), identifying semi-major and semi-minor axes.

Both shapes intersect where their equations are equal, defining the limits of our integral. The radial limit derived from the ellipse is more complex, represented by \(r = \frac{2}{\sqrt{\cos^2 \theta + 2\sin^2 \theta}}\), reflecting its asymmetrical stretch compared to the circle.

Trigonometric Identities

Trigonometric identities often appear in integrals involving polar coordinates. They simplify expressions and help evaluate integrals by revealing simpler forms or symmetry.

• A crucial identity used in this exercise is \(\cos^2 \theta = \frac{1 + \cos 2\theta}{2}\). It breaks down squared trigonometric terms into linear expressions, making integration much easier.
• Another identity derived from the first is \(\cos^4 \theta = \left( \frac{1 + \cos 2\theta}{2} \right)^2\), which is used to simplify quartic trigonometric terms to facilitate integration.

These identities are useful for the integration part of the problem, transforming complex powers of \(\cos \theta\) into simpler forms, reducing the complexity of the final integration from a few intricate terms to basic integrable functions. Employing such identities requires not just memorization but also understanding their derivations, as this helps in recognizing where and how they can be applied most effectively.