Problem 23 - Essential Calculus

Question

In Exercises 23-26, use a graphing utility to graph the polar equations and find the area of the given region. Common interior of \(r=3-2 \sin \theta\) and \(r=-3+2 \sin \theta\)

Step-by-Step Solution

Answer

The combined area under the two curves \(r=3-2 \sin \theta\) and \(r=-3+2 \sin \theta\) between \(\theta = \frac{\pi}{2}\) and \(\theta = \frac{3\pi}{2}\) can be calculated by the steps above. The final area calculation will require evaluation of these steps, which is beyond the scope of this response.

Step 1: Graphing the Known Functions

Using a graphing tool, draw the graphs of \(r=3-2 \sin \theta\) and \(r=-3+2 \sin \theta\). Observe where they intersect. These intersection points will be used as the limits of integration.

Step 2: Finding Intersection Points

To find the points where the graphs intersect, the equations \(3-2 \sin \theta = -3+2 \sin \theta\) is solved. This can be simplified to give \(\sin \theta = 1\) and \(\sin \theta = -1\), which correspond to \(\theta = \frac{\pi}{2}\) and \(\theta = \frac{3\pi}{2}\) respectively. Therefore, \(\theta = \frac{\pi}{2}\) and \(\theta = \frac{3\pi}{2}\) are the limits of integration.

Step 3: Computing the Area

To find the area of the region bounded by the two curves, the area under each curve is calculated separately from \(\frac{\pi}{2}\) to \(\frac{3\pi}{2}\) and then the two results are added together. After finding the integrals, evaluate them at \(\frac{3\pi}{2}\) and \(\frac{\pi}{2}\) and subtract the results to calculate the area: \(A = A_{1} + A_{2}\)

Key Concepts

Graphing Polar Equations

Graphing polar equations can be an intriguing journey since, unlike Cartesian coordinates, polar coordinates involve both a radius and an angle. In our original problem, we have two polar equations: \(r=3-2\sin\theta\) and \(r=-3+2\sin\theta\). By converting these into a visual form, we gain insight into their unique behavior.
You can use a graphing utility to draw these curves. This helps in visualizing them in the polar coordinate plane. Here, the variable \(r\) represents the distance from the origin, also known as the pole, while \(\theta\) is the angle measured in radians from the positive x-axis (polar axis).
As you plot the graphs, notice how the sinusoidal component alters the shape. Polar curves can form circles, cardioids, or even more complex shapes depending on the equation.
In this problem, these equations will intersect at certain points, providing us with crucial information about their limits for further calculations, especially for finding the area of polar regions.

Area of Polar Regions

Calculating the area of a region defined by polar curves involves using definite integrals. It requires setting integration limits that match the range of angles over which the region is defined. For the problem at hand, once we determine the intersection points of the curves, these act as limits of integration. In this case, the curves intersect at \(\theta = \frac{\pi}{2}\) and \(\theta = \frac{3\pi}{2}\).
To compute the area, you will apply the formula for finding the area enclosed by a polar curve:

• \(A = \frac{1}{2} \int_{a}^{b} (r^2) \, d\theta\)

where \(r\) represents the given polar equation, and \([a, b]\) are the limits of integration.

• For \(r=3-2\sin\theta\), evaluate the integral over \(\theta = \frac{\pi}{2}\) to \(\theta = \frac{3\pi}{2}\).
• Similarly, for \(r=-3+2\sin\theta\), compute separately over the same limits.
• Combine these areas by adding the results from the integrals.

The correct setup and execution of these steps yield the area of the shared interior between the two curves.

Intersection of Polar Curves

Finding the intersection points of polar curves is pivotal as it sets the stage for area calculations among other analyses. In our case, intersections occur when both polar equations yield the same radius \(r\) for a particular angle \(\theta\).
By equating the two given polar equations, \(3-2\sin\theta = -3+2\sin\theta\), we deduce the sin value and solve for \(\theta\).
Solving this effectively provides the angles at which the two curves cross each other.

• Solve \(2\sin\theta = 6\) to find \(\sin\theta = 1\).
• This leads us to \(\theta = \frac{\pi}{2}\).
• Identifying that \(\sin\theta = -1\), results in \(\theta = \frac{3\pi}{2}\).

These \(\theta\) values give us the critical points where the curves intersect, framing our limits for further integration in the given problem. Understanding these intersections enriches your grasp of how polar equations behave and interact on a graph.