Problem 3 - Advanced Engineering Mathematics

Question

In Problems, use a double integral in polar coordinates to find the area of the region bounded by the graphs of the given polar equations. $$ \begin{array}{lll} \boldsymbol{r} & 2 \sin \theta, \boldsymbol{r} & 1, \text { common area } \end{array} $$

Step-by-Step Solution

Answer

The area of the common region is \( \frac{\pi}{3} - \frac{\sqrt{3}}{4} \).

Step 1: Identify and Sketch the Region

First, identify the polar equations. The first equation is \( r = 2 \sin \theta \), which is a circle centered at \((0,1)\) with radius 1. The second equation \( r = 1 \) is a circle centered at the origin with radius 1. Sketch these graphs to visualize the regions they bound.

Step 2: Determine Intersection Points

To find the region common to both curves, we need to find their points of intersection. Setting \( 2 \sin \theta = 1 \), solve for \( \theta \). This equation gives \( \sin \theta = \frac{1}{2} \), so \( \theta = \frac{\pi}{6} \) and \( \theta = \frac{5\pi}{6} \). These are the angles where the curves intersect.

Step 3: Set Up the Double Integral

The area of the region common to both curves is found using the double integral in polar coordinates: \[A = \int_{\theta_1}^{\theta_2} \int_{r_1(\theta)}^{r_2(\theta)} r \, dr \, d\theta\]Substitute the limits from \(\theta = \frac{\pi}{6}\) to \(\theta = \frac{5\pi}{6}\), with inner integral limits from the inner curve \(r = 2\sin\theta\) to the outer curve \(r = 1\).

Step 4: Compute the Inner Integral with respect to r

The inner integral with respect to \( r \) is given by:\[\int_{2\sin\theta}^{1} r \, dr = \left[ \frac{r^2}{2} \right]_{2\sin\theta}^{1}\]Evaluate this expression to get:\[\frac{1^2}{2} - \frac{(2\sin\theta)^2}{2} = \frac{1}{2} - 2\sin^2\theta\]

Step 5: Compute the Outer Integral with respect to θ

With the result from the inner integral, compute the outer integral:\[\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \left( \frac{1}{2} - 2\sin^2\theta \right) \, d\theta\]Use the identity \( \sin^2\theta = \frac{1 - \cos(2\theta)}{2} \) to simplify:\[= \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \left( \frac{1}{2} - \left( 1 - \cos(2\theta) \right) \right) \, d\theta = \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \left( -\frac{1}{2} + \cos(2\theta) \right) \, d\theta\]

Step 6: Evaluate the Outer Integral

Evaluate each part separately:\[\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \left( -\frac{1}{2} \right) \, d\theta = -\frac{1}{2} \times \left( \frac{5\pi}{6} - \frac{\pi}{6} \right)\]And \[\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \cos(2\theta) \, d\theta = \left[ \frac{1}{2} \sin(2\theta) \right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\]Evaluate these integrals and combine results.

Step 7: Calculate the Final Answer

After integrating, combining the results gives:\[A = -\frac{1}{2} \left( \frac{4\pi}{6} \right) + \frac{1}{2} \left( \sin\left(\frac{5\pi}{3}\right) - \sin\left(\frac{\pi}{3}\right) \right)\]Simplify further to find the final numerical value for the area of the region.

Key Concepts

Polar Equations

Polar equations help us describe curves on a polar coordinate plane. In a polar coordinate system, each point is determined by an angle and a distance from the origin. The notation \( r = f(\theta) \) defines curves in this system.

• Here, \( r \) is the radius or distance from the pole (origin) to the curve.
• \( \theta \) is the angle measured from the polar axis.

For example, the equation \( r = 2 \sin \theta \) represents a circle in polar coordinates, centered at the point \((0,1)\) with a radius of 1. The nature of polar equations often highlights symmetries which are not as obvious in rectangular coordinates.

Intersection Points

Intersection points are crucial to defining the bounds of a region. They tell us where curves meet or intersect, providing limits for integration. To find intersection points in polar coordinates, set the polar equations equal to each other and solve for \( \theta \). For example, given the equations \( r = 2 \sin \theta \) and \( r = 1 \):

1. Set \( 2 \sin \theta = 1 \).
2. Solve for \( \sin \theta = \frac{1}{2} \).
3. Find \( \theta = \frac{\pi}{6} \) and \( \theta = \frac{5\pi}{6} \).

These angles are where the two curves intersect, meaning at those angles, both curves have the same radius, forming the boundaries of the region of interest.

Area of a Region

The area of a region between two polar curves can be found through a double integral in polar coordinates. This method involves integrating over the region to calculate the enclosed area between two curves. The formula is:

\[ A = \int_{\theta_1}^{\theta_2} \int_{r_1(\theta)}^{r_2(\theta)} r \, dr \, d\theta \]

Here,

• \( \theta_1 \) and \( \theta_2 \) are the intersection points found previously, marking the angular bounds of the region.
• \( r_1(\theta) \) and \( r_2(\theta) \) provide the radial limits; typically, these are the two intersecting curves' equations.

This integration approach directly calculates the region's area, defined by the circle arcs and the angles where they intersect.

Integration Techniques

Integration techniques in this context involve handling double integrals in polar coordinates. The integral must be computed step-by-step:

• First, calculate the inner integral with respect to \( r \). This involves integrating \( r \) from the inner circle \( r = 2 \sin \theta \) to the outer circle \( r = 1 \). For instance: \[ \int_{2\sin\theta}^{1} r \, dr = \left[ \frac{r^2}{2} \right]_{2\sin\theta}^{1} \]
• Next, compute the result of this inner integral.

Then, the outer integral is tackled with respect to \( \theta \), considering the resulting expression from the inner integration. Correctly evaluating these integrals will result in finding the area.

Using identities like \( \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \) simplifies trigonometric terms, aiding in integration. These simplifications are essential for calculating regions bounded by curves, especially when integrating complex equations.