Problem 28 - Calculus Early Transcendentals

Question

\(23-28\) Find the area of the region that lies inside the first curve and outside the second curve. $$ r=3 \sin \theta, \quad r=2-\sin \theta $$

Step-by-Step Solution

Answer

The area is evaluated to be approximately 1.179 units.

Step 1: Understand the Problem

We need to find the area of the region that is inside the curve \( r=3\sin\theta \) and outside the curve \( r=2-\sin\theta \). This involves finding the intersection points of these two curves and then determining the bounds and area of the region.

Step 2: Find Intersection Points

To find the intersection points of the curves \( r=3\sin\theta \) and \( r=2-\sin\theta \), set the equations equal to each other: \( 3\sin\theta = 2-\sin\theta \). Solving for \( \sin\theta \), we get \( 4\sin\theta = 2 \) or \( \sin\theta = \frac{1}{2} \). The solutions for \( \theta \) are \( \frac{\pi}{6} \) and \( \frac{5\pi}{6} \).

Step 3: Set Up the Integral for Area

The area inside \( r=3\sin\theta \) and outside \( r=2-\sin\theta \) for \( \theta \) from \( \frac{\pi}{6} \) to \( \frac{5\pi}{6} \) is given by the integral: \[ A = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \left( (3\sin\theta)^2 - (2-\sin\theta)^2 \right) d\theta. \]

Step 4: Simplify the Integral

Simplify the integrand: \( 9\sin^2\theta - (4 - 4\sin\theta + \sin^2\theta) = 8\sin^2\theta + 4\sin\theta - 4 \). The integral becomes: \[ A = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} (8\sin^2\theta + 4\sin\theta - 4) d\theta. \]

Step 5: Evaluate the Integral

To evaluate the integral, consider the terms separately and use trigonometric identities if needed. Use the identity \( \sin^2\theta = \frac{1 - \cos 2\theta}{2} \) to integrate \( \sin^2\theta \). Solve \[ \int \sin^2 \theta \, d\theta, \int \sin \theta \, d\theta, \text{ and } \int \, d\theta \] individually and compute the values between the limits \( \frac{\pi}{6} \) to \( \frac{5\pi}{6} \). Combine these results to get the final area.

Step 6: Combine Results

Combine the evaluated integrals from Step 5 to find the total area. Compute each part and then sum/stretch them to find the area between the two curves for the given range.

Key Concepts

Area between curves

To find the area between two polar curves, it's crucial to understand the concept of boundaries within the coordinate plane. In polar coordinates, the radial distance from the origin is given by a function of the angle, \(\theta\). The area between two curves, say \(r=f(\theta)\) and \(r=g(\theta)\), is determined by integrating the square of these functions over the region where they intersect.
This can be imagined as subtracting the smaller portion from the larger one over a particular range of \(\theta\).

• First, identify where the curves meet, which involves finding their intersection points.
• Next, set up a definite integral to compute the area by squaring and subtracting these radius values.

This way, you capture the area that lies inside one polar plot and outside another.

Trigonometric identities

Trigonometric identities are mathematical equations that involve trigonometric functions like sine, cosine, and tangent. They are crucial in simplifying expressions and solving integrals in polar coordinates.
For example, to simplify the term \(\sin^2\theta\), you use the identity:

• \(\sin^2\theta = \frac{1 - \cos(2\theta)}{2}\)

This identity transforms a quadratic trigonometric term into a simpler linear form that is easier to integrate. Another important identity used in polar integration is the basic integral of sine:

• \(\int \sin \theta \, d\theta = -\cos \theta + C\)

These identities help simplify complex integrals, making calculations more manageable.

Definite integrals

Definite integrals give the net area between a curve and the horizontal axis over a specific interval, providing the net sum of calculated values.
In polar coordinates, definite integrals are used to calculate areas by integrating a radial function \(r\) squared. Thus, the integral for the area is: \[ \frac{1}{2} \int_{\theta_1}^{\theta_2} \left( r_{\text{outer}}^2 - r_{\text{inner}}^2 \right)\,d\theta \]The limits of this integral are the angles where the curves intersect, as these define the bounds of the region of interest.

• The \(\frac{1}{2}\) factor comes from the polar area element formula.
• The integrands, \(r_{\text{outer}}^2\) and \(r_{\text{inner}}^2\), describe the square of the curve's radius at each angle \(\theta\).

This integral setup captures the true area bounded by these curves.

Intersection points

Finding intersection points of polar curves is a fundamental step when calculating the area between them. Intersection points are the specific \(\theta\) values where two polar equations have the same radius \(r\).
To determine these points, equate the polar equations:

• For example, if you have \(r=3\sin\theta\) and \(r=2-\sin\theta\), set them equal: \(3\sin\theta = 2 - \sin\theta\).

Solve for \(\theta\) to find where the equations coincide, in terms of angle:

• This gives \(\sin\theta = \frac{1}{2}\), which corresponds to angles \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{5\pi}{6}\).

These points of intersection define the boundaries for integrating area and signal where the curves intersect each other on the polar plane.