Problem 3 - Essential Calculus

Question

In Exercises 3 and 4, find the area of the region bounded by the graph of the polar equation using (a) a geometric formula and (b) integration. $$ r=8 \sin \theta $$

Step-by-Step Solution

Answer

The area of the region bounded by the graph of the polar equation \( r = 8 \sin \theta \) is \(16 \pi\) square units.

Step 1: Solution using geometric formula

The given polar equation \( r = 8 \sin \theta \) represents a circle with radius half of the fixed number (which is 8 in this case). Hence the radius of the circle is \(r = 4\). The area \(A\) of a circle with radius \(r\) is given by the formula \(A = \pi r^2\). Substituting \(r = 4\) into the formula gives \(A = \pi (4)^2\)

Step 2: Solution using integration.

The area \(A\) for a polar curve is given by the formula \( A = \frac{1}{2} \int_{a}^{b} r^2 d \theta \). For a full circle in polar form, \( a = 0, b = 2 \pi \). Substituting \( r = 8 \sin \theta \) into the formula gives \( A = \frac{1}{2} \int_{0}^{2 \pi} (8 \sin \theta)^2 d\theta \). Simplify the integral to \( 32 \int_{0}^{2 \pi} \sin^2 \theta d\theta \). Using the power-reduction identity \( \sin^2\theta = \frac{1 - \cos 2\theta }{2}\), replace \(\sin^2\theta\) and simplify.

Step 3: Evaluation of integral

The integral becomes \( A = 32 \int_{0}^{2\pi} \frac{1 - \cos 2 \theta}{2} d\theta = 16 \pi - 16 \int_{0}^{2\pi} \cos 2\theta d\theta\). The second integral evaluates to zero as it involves a complete cycle of the cosine function. Thus, \(A = 16 \pi\)

Key Concepts

Geometric Formula

When working with polar equations, often the shape described by the equation might be a familiar geometric figure, like a circle. In this exercise, the polar equation \( r = 8 \sin \theta \) defines such a figure. The equation is a classic representation of a circle in polar coordinates. Here, the figure described by the equation is a circle with a maximum radius, also known as the diameter in Cartesian terms.
To determine the actual area using the geometric formula, we realize that the amplitude \( 8 \) corresponds to the circle's diameter. Thus, the radius of the circle is half of that, which is \( r = 4 \).
The formula for the area \( A \) of a circle is \( A = \pi r^2 \). Substituting \( r = 4 \) gives \( A = \pi (4)^2 = 16\pi \). This method is straightforward because it doesn't require calculus, but deep understanding of polar graphs is essential to spot such cases.

Integration in Polar Coordinates

Integration in polar coordinates allows us to solve problems for regions and shapes that are more naturally described with polar equations rather than Cartesian coordinates. The key formula for finding the area \( A \) of a region enclosed by a polar curve \( r(\theta) \) from \( \theta = a \) to \( \theta = b \) is:

\[ A = \frac{1}{2} \int_{a}^{b} r^2 \, d\theta \]

The exercise requires finding the area enclosed by the circle described by \( r = 8 \sin \theta \). Plugging into the formula, we consider a full revolution where \( \theta \) varies from 0 to \( 2\pi \). Replacing \( r = 8 \sin \theta \) in the integral gives
\[ A = \frac{1}{2} \int_{0}^{2\pi} (8 \sin \theta)^2 \, d\theta \]
Simplifying, the integral becomes \( 32 \int_{0}^{2\pi} \sin^2 \theta \, d\theta \). When integrating in polar coordinates, understanding and applying trigonometric identities becomes crucial to solve the resulting integrals easily.

Power-Reduction Identity

The power-reduction identity is an essential trigonometric identity that simplifies expressions involving squares of sine or cosine functions. Specifically, \( \sin^2 \theta \) is simplified using the identity:

\[ \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \]

This transformation is vital for evaluating integrals that result from polar area calculations because it turns a squared trigonometric function into a sum of simpler trigonometric functions. In the context of our problem, replacing \( \sin^2 \theta \) with \( \frac{1 - \cos 2\theta}{2} \) helps reduce the complexity of the integral and eventually simplifies further calculation. The integral then becomes:

\[ A = 32 \int_{0}^{2\pi} \frac{1 - \cos 2\theta}{2} \, d\theta \]

This further evaluates to \( 16\pi \), considering the integral of \( \cos 2\theta \) over a full period is zero. Thus, the power-reduction identity streamlines the integration process substantially and leads to the correct area calculation in polar coordinates.