Problem 5 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the graph of the given equation and find the area of the region bounded by it. $$ r=3-3 \sin \theta $$

Step-by-Step Solution

Answer

The area of the bounded region is \(20.25\pi\).

Step 1: Identify the Type of Graph

The equation given is in polar form, which means we are dealing with a polar graph. Specifically, the equation \(r = 3 - 3 \sin \theta\) is a limaçon. Due to the subtraction of the sinusoidal component, this specific graph will be a cardioid with a 'dimple.'

Step 2: Sketch the Graph

To sketch the graph, you can convert the polar equation to Cartesian coordinates or plot key points directly from the polar form. Note that the cardioid will have a maximum radius of 6 at \(\theta = \frac{3\pi}{2}\) and crosses the pole at \(\theta = 0\), creating a looping heart-like shape.

Step 3: Find the Limits for Integration

To find the area, identify the bounds of \(\theta\) for one full rotation. For a cardioid, this typically ranges from \(\theta = 0\) to \(\theta = 2\pi\).

Step 4: Set Up the Integral for Area

The area enclosed by a polar graph is calculated using the integral \( A = \frac{1}{2} \int_{a}^{b} r^2\,d\theta \). Substituting \(r = 3 - 3 \sin \theta\), the formula becomes \( A = \frac{1}{2} \int_{0}^{2\pi} (3 - 3 \sin \theta)^2\,d\theta \).

Step 5: Simplify the Integral

Expand the squared term: \((3 - 3\sin\theta)^2 = 9 - 18\sin\theta + 9\sin^2\theta\). Substitute this into the integral, resulting in \( \frac{1}{2} \int_{0}^{2\pi} (9 - 18\sin\theta + 9\sin^2\theta)\,d\theta \).

Step 6: Evaluate the Integral

Break it into separate integrals: \( \frac{1}{2} \left(9\int_{0}^{2\pi} d\theta - 18\int_{0}^{2\pi} \sin\theta\,d\theta + 9\int_{0}^{2\pi} \sin^2\theta\,d\theta \right) \). The first integral evaluates to \(18\pi\), the second to \(0\), and the third to \(\frac{9}{2}\pi\) using the identity \( \sin^2\theta = \frac{1}{2}(1 - \cos 2\theta) \).

Step 7: Sum and Calculate the Area

Combine the evaluated parts: \( \frac{1}{2} \left(18\pi + \frac{9}{2}\pi\right) = \frac{1}{2}(40.5\pi) = 20.25\pi \). Thus, the area of the cardioid bounded region is \(20.25\pi\).

Key Concepts

Limaçon

The term "limaçon" refers to a family of curves in polar coordinates, characterized by equations of the form \(r = a + b \sin \theta\) or \(r = a + b \cos \theta\). Depending on the relationship between \(a\) and \(b\), the limaçon can appear in several distinct shapes.

• If \(a = b\), the limaçon takes on a cardioid shape, which resembles a heart.
• If \(a > b\), it develops a dimple but doesn't loop back on itself.
• When \(b > a\), the curve crosses itself, creating an inner loop.

The equation \(r = 3 - 3 \sin \theta\) clearly points to a special case of a limaçon, known as a cardioid, which we will explore further in the next section.

Cardioid

A cardioid is a special type of limaçon that occurs when \(a = b\). The equation \(r = 3 - 3 \sin \theta\) is a cardioid because the coefficients of both terms are equal.

• This results in a distinctive 'heart-shaped' curve.
• The term "cardioid" comes from the Greek word "kardia," meaning heart.

For our specific cardioid equation, the shape is formed symmetrically with respect to the vertical axis. The maximum radius occurs when \(\theta = \frac{3\pi}{2}\), which is when the sine function reaches its minimum, giving the curve a length of 6 units along the negative vertical axis.
At \(\theta = 0\), the cardioid dips to the pole, forming the characteristic "dimple," indicative of the cardioid's nature.

Integration in Polar Coordinates

Integration in polar coordinates is a method to calculate areas for curves that are best described using polar equations. The formula to find the area enclosed by a polar curve \(r = f(\theta)\) is \[ A = \frac{1}{2} \int_{a}^{b} r^2 \, d\theta \].

• The \(\frac{1}{2}\) factor accounts for the radial nature of polar systems.
• The integral bounds \(a\) and \(b\) typically represent one full cycle of the curve.

In our problem, \(r = 3 - 3 \sin \theta\), and we integrate from \(0\) to \(2\pi\) to cover the entire cardioid. This step consolidates the symmetry and complexity of polar graphs into a single integral expression, simplifying computation of areas for polar graphs.

Area Calculation

Calculating the area enclosed by a polar curve involves evaluating the integral \(A = \frac{1}{2} \int_{a}^{b} (3 - 3 \sin \theta)^2 \, d\theta\).

• First, expand \((3 - 3\sin\theta)^2\) to get \(9 - 18\sin\theta + 9\sin^2\theta\).
• Break this into separate integrals for each term.

Evaluate each component:

• The first term, \(9 \int_{0}^{2\pi} d\theta\), results in \(18\pi\).
• The second term, \(-18 \int_{0}^{2\pi} \sin\theta \, d\theta\), evaluates to 0.
• The third term, \(9 \int_{0}^{2\pi} \sin^2\theta \, d\theta\), simplifies using the identity \( \sin^2\theta = \frac{1}{2}(1 - \cos 2\theta)\), resulting in \(\frac{9}{2}\pi\).

Summing these areas gives \(\frac{1}{2} (18\pi + \frac{9}{2}\pi) = 20.25\pi\). Thus, the area of the cardioid's bounded region is \(20.25\pi\). This approach of breaking down the integral allows for straightforward calculations in polar systems.