Problem 42 - Calculus Early Transcendentals

Question

In Exercises \(37-40,\) you found the intersection points of pairs of curves. Find the area of the entire region that lies within both of the following pairs of curves. $$r=2+2 \sin \theta \text { and } r=2-2 \sin \theta$$

Step-by-Step Solution

Answer

Question: Find the area of the region that lies within the polar curves $$r=2+2 \sin \theta \text{ and } r=2-2 \sin \theta$$. Answer: The area of the entire region that lies within the polar curves is 16 square units.

Step 1: Find the points of intersection

To find the intersection points between these two curves, we set their equations equal to each other and solve for \(\theta\): $$2+2 \sin \theta = 2-2 \sin \theta$$ Solving for \(\sin \theta\), we get: $$4 \sin \theta = 0$$ $$\sin \theta = 0$$ The solutions for \(\theta\) are: $$\theta = 0 \text{ and } \theta = \pi$$.

Step 2: Determine limits of integration

Since the curves intersect at \(\theta = 0\) and \(\theta = \pi\), these values will be the limits of our integral. We need to decide which curve is the "outer curve" and which one is the "inner curve" to apply our area formula properly. Looking at the equations, we can see that \(r=2+2 \sin \theta\) is greater than \(r=2-2 \sin \theta\) for all \(\theta\) between \(0\) and \(\pi\), so we will use this information in the next step to determine the integrals.

Step 3: Calculate the enclosed area

Now, we can use the polar area formula to calculate the enclosed area between these curves. The area formula is: $$A = \frac{1}{2} \int_{\alpha}^{\beta} (r_{\text{outer}}^2 - r_{\text{inner}}^2) d\theta$$ In our case, \(r_{\text{outer}} = 2+2 \sin \theta\), \(r_{\text{inner}} = 2-2 \sin \theta\), \(\alpha = 0\), and \(\beta = \pi\). Plugging in the values, we get: $$A = \frac{1}{2} \int_{0}^{\pi} [(2+2\sin\theta)^2 - (2-2\sin\theta)^2] d\theta$$

Step 4: Simplify and find the integral

Let's simplify the integrand and find the integral: $$A = \frac{1}{2} \int_{0}^{\pi} (4 + 8\sin\theta + 4\sin^2\theta - 4 + 8\sin\theta - 4\sin^2\theta) d\theta$$ $$A = \frac{1}{2} \int_{0}^{\pi} (16\sin\theta) d\theta$$ $$A = 8 \left[-\cos\theta\right]_0^\pi$$ $$A = 8(-(-1) - (-1))$$ $$A = 16$$ The area of the entire region that lies within the polar curves $$r=2+2 \sin \theta \text{ and } r=2-2 \sin \theta$$ is 16 square units.

Key Concepts

Intersection of Curves

When working with polar coordinates, finding the intersection of curves is an important skill. For polar equations, the process involves setting the given equations equal to each other. You solve for the angle \( \theta \) where the radial distances \( r \) are the same on both curves. In the exercise, we have two equations: \( r = 2 + 2\sin\theta \) and \( r = 2 - 2\sin\theta \).
By equating them, we set:

• \( 2 + 2\sin\theta = 2 - 2\sin\theta \)

Solving for \( \sin\theta \) reveals that \( \sin\theta = 0 \). This condition is met when \( \theta = 0 \) or \( \theta = \pi \). These intersections occur at the angles where the curves meet. Understanding this lets you properly define your boundary limits for the area calculation.

Polar Area Formula

The polar area formula is a powerful tool for finding areas between polar curves. It allows you to compute the area of a sector in a polar coordinate system, which is different from a Cartesian plane. The formula is generally given by:

• \( A = \frac{1}{2} \int_{\alpha}^{\beta} (r_{\text{outer}}^2 - r_{\text{inner}}^2) \, d\theta \)

The integral is calculated between two angular limits \( \alpha \) and \( \beta \), which are the intersection points of the curves. In the exercise, since we found \( \theta = 0 \) and \( \theta = \pi \) as points of intersection, these are the limits used for the integration. It is crucial to ensure that \( r_{\text{outer}} \) represents the curve with a larger \( r \) value within this interval compared to \( r_{\text{inner}} \), which is the curve with the smaller \( r \) value. This arrangement ensures the correct calculation of area.

Integration in Polar Coordinates

Integration in polar coordinates involves evaluating an integral to find areas bound by curves. It's a method specifically adapted for dealing with circular and radial symmetries that occur naturally in polar systems. In our problem:

• We identify that \( r_{\text{outer}} = 2 + 2\sin\theta \)
• And \( r_{\text{inner}} = 2 - 2\sin\theta \)

As mentioned, the integration limits are \( \theta = 0 \) to \( \theta = \pi \). The integration step involves simplifying the expression inside the integral:

• Simplifying: \( [(2+2\sin\theta)^2 - (2-2\sin\theta)^2] \)
• Which resolves to \( 16\sin\theta \)

By integrating \( 16\sin\theta \) over \( \theta \) from \( 0 \) to \( \pi \), the resultant area of \( 16 \) square units represents the entire region enclosed by both curves. This conceptual understanding highlights the elegance of polar integration in calculating areas defined by curved boundaries.