Problem 25 - Essential Calculus

Question

Use a graphing utility to graph the polar equations and find the area of the given region. Common interior of \(r=4 \sin \theta\) and \(r=2\)

Step-by-Step Solution

Answer

The area of the common interior of the graphs of the given polar equations is \(2\pi\) square units.

Step 1: Graph the Equations

Start by graphing the polar equations using a graphing utility. The graph of \(r=4 \sin \theta\) is a circle with radius 2 centered at (0,2) while that of \(r=2\) is a circle with radius 2 centered at the origin.

Step 2: Find the Intersection Points

Next, determine the points where the graphs intersect. These are the values of \(\theta\) for which the two equations are equal, i.e. when \(4 \sin \theta = 2\). Solve this to get \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{5\pi}{6}\).

Step 3: Calculate the Area

Now calculate the area of the common interior. This is given by \(\frac{1}{2}\) multiplied by the integral from \(\theta_1\) to \(\theta_2\) of \((f(\theta))^2 - (g(\theta))^2 d\theta\), where f and g are the given polar equations. In this case, it would be \(\frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} ((4 \sin \theta)^2 - 2^2)d\theta \). Solve the integral to find the area.

Key Concepts

Polar Equations

Polar coordinates provide a unique way to describe points on a plane. Unlike Cartesian coordinates which use x and y axes, polar coordinates use a distance from the origin, known as radius (r), and an angle (θ) from the positive x-axis. Polar equations, like the ones used in our exercise, often graph curves that are circular or have rotational symmetry. Two common forms of polar equations are circles and lemniscates, each defined by its specific equation. For example:

• The equation \( r=4 \sin \theta \) represents a circle centered at (0,2) with radius 2 when graphed.
• The equation \( r=2 \) represents a circle with a center at the origin, also of radius 2.

Intersection Points

Finding intersection points in polar coordinates involves setting the two polar equations equal to each other, as these points occur where both equations have the same radius for the same angle.For the equations \( r=4 \sin \theta \) and \( r=2 \), intersection points are where the distance from the origin, r, overlaps for both graphs.

Setting these equal gives us \( 4 \sin \theta = 2 \), from which we solve to find \( \theta = \frac{\pi}{6} \) and \( \theta = \frac{5\pi}{6} \).

This means the graphs of the circles intersect at these specific angles, forming the boundary for calculating the area of the overlapping region.

Graphing Utility

A graphing utility is a versatile tool for students and mathematicians to visualize polar equations. It's a software application that can graph complex mathematical equations and show intersections, allowing for easy examination of these points.To graph polar equations like \( r=4 \sin \theta \) and \( r=2 \), input each equation into the graphing utility.

• Ensure that you adjust the settings to polar mode, as this is essential to view how these equations map onto the polar plane.
• Observe the manner in which the circles overlap by following the plotted graph.

Utilizing a graphing utility greatly enhances understanding by giving a visual context to theoretical calculations, making it simpler to find exact points of intersection and visualize the area in question.

Integral Calculation

When calculating the area between intersecting polar graphs, integration becomes your best tool. The integral makes it possible to determine the exact area of the region enclosed by two polar curves.To find this area, the formula involves an integration technique:

• The formula is given by \( \frac{1}{2} \int_{\theta_1}^{\theta_2} ((f(\theta))^2 - (g(\theta))^2)d\theta \), where \( f(\theta) \) and \( g(\theta) \) are our polar equations.
• The limits of integration, \( \theta_1 \) and \( \theta_2 \), are the angles found at the points of intersection.

For our equations, this translates to \( \frac{1}{2} \) times the integral from \( \frac{\pi}{6} \) to \( \frac{5\pi}{6} \) of \( ((4 \sin \theta)^2 - 2^2)d\theta \).
Solving this integral gives the area of the region common to both polar graphs, combining mathematical precision with visual intuition.