Problem 47 - Calculus Early Transcendentals

Question

Find the areas of the following regions. The region common to the circles \(r=2 \sin \theta\) and \(r=1\)

Step-by-Step Solution

Answer

Answer: \(\frac{4\pi}{3}\).

Step 1: Find the Intersection Points

To find the intersection points, we will set the equations \(r=2\sin\theta\) and \(r=1\) equal to each other: \(2\sin\theta = 1\) Now solve for \(\theta\): \(\sin\theta = \frac{1}{2}\) \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}\) The intersection points are at \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{5\pi}{6}\).

Step 2: Calculate the Area of Each Region

To find the areas of each region, we will use the polar coordinates area formula: \(A = \frac{1}{2} \int_\alpha^\beta r^2 d\theta\) Firstly, calculate the area enclosed by the circle \(r=2\sin\theta\): \(A_1 = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} (2\sin\theta)^2 d\theta\) \(A_1 = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} 4\sin^2\theta d\theta\) Next, calculate the area enclosed by the circle \(r=1\): \(A_2 = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} 1^2 d\theta\) \(A_2 = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} d\theta\) Now, we need to find the area of the common region.

Step 3: Find the Area of the Common Region

To find the area of the common region, we need to subtract the smaller circle's area (\(A_2\)) from the larger circle's area (\(A_1\)): \(A_{common} = A_1 - A_2\) \(A_{common} = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} 4\sin^2\theta d\theta - \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} d\theta\) Now, evaluate the integrals: \(A_{common} = \frac{1}{2} [2\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} (2\sin^2\theta - 1) d\theta]\) Perform the integration: \(A_{common} = \frac{1}{2} [2(\frac{1}{2}[-\cos(2\theta)]_{\frac{\pi}{6}}^{\frac{5\pi}{6}} + \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} d\theta)]\) \(A_{common} = \frac{1}{2} [2(-\frac{1}{2}[\cos(2\cdot \frac{5\pi}{6})-\cos(2\cdot \frac{\pi}{6})] + [\theta]_{\frac{\pi}{6}}^{\frac{5\pi}{6}})]\) Calculate the result: \(A_{common} = \frac{1}{2} [2(-\frac{1}{2}[-\sqrt{3}+\sqrt{3}] + [2\pi-\frac{\pi}{3}])]\) \(A_{common} = \frac{1}{2} [2\cdot\frac{4\pi}{3}]\) \(A_{common} = \frac{4\pi}{3}\) So, the area of the common region is \(\frac{4\pi}{3}\).

Key Concepts

Intersection Points

When dealing with polar coordinates, finding the intersection points of two curves helps in understanding how they overlap. For the given problem, the equations are given as the circles: \(r = 2 \sin \theta\) and \(r = 1\). To find the intersection points, you equate the two equations to find the common values of \(\theta\). Let's break it down:

• Set the equations equal: \(2 \sin \theta = 1\).
• Solve for \(\theta\): you get \(\sin \theta = \frac{1}{2}\).
• Determine the values of \(\theta\): these are \(\theta = \frac{\pi}{6}\) and \(\frac{5\pi}{6}\).

These angles correspond to the points where the two circles intersect. Such points are key to setting boundaries for integrations when calculating areas in polar coordinates.

Polar Area Formula

The polar area formula is crucial to finding the area enclosed by curves represented in polar coordinates. The formula reads: \[ A = \frac{1}{2} \int_\alpha^\beta r^2 \, d\theta \]This formula calculates the area 'swept out' by the polar curve from angle \(\alpha\) to \(\beta\). For our problem:1. **For circle \(r = 2 \sin \theta\):** - Insert into the formula: \(A_1 = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} (2\sin\theta)^2 \, d\theta\). - Simplify: it becomes \(\frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} 4\sin^2\theta \, d\theta\).2. **For circle \(r = 1\):** - Form the setup: \(A_2 = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} 1^2 \, d\theta\).These steps illustrate how using polar coordinates, you can focus on calculating areas defined by circular arcs. The calculations involve finding squared functions of the radius and integrating within the defined limits.

Trigonometric Integration

Trigonometric integration is a technique essential to calculating areas in polar coordinates. In the example provided, this technique was used to determine the area of the region common to the two circles. Here's how it works:- **Expressing functions using trigonometric identities:** - For \(4\sin^2\theta\), apply the identity \(\sin^2\theta = \frac{1 - \cos(2\theta)}{2}\). - The integral becomes a bit simpler: \(\int 4\sin^2\theta \, d\theta = 2 \int (1 - \cos(2\theta)) \, d\theta\).- **Breaking down each integral:** - Split the integration into two parts: \(\int 1 \, d\theta\) and \(\int \cos(2\theta) \, d\theta\).Through this detailed approach, evaluating integrals involving trigonometric functions becomes manageable. This integration method enables the calculation of complex shapes defined in polar form by reducing them to elementary integrals. Thus, you can find the exact area of regions like the common section of overlapping circles.