Problem 201 - Calculus Volume 3

Question

Find the area of the described region. Enclosed by \(r=6 \sin \theta\)

Step-by-Step Solution

Answer

The area is \(9\pi\).

Step 1: Understand the Polar Equation

The given equation is in polar coordinates: \(r = 6 \sin \theta\). This represents a circle with its center at \((3, \frac{\pi}{2})\) and radius 3. In polar form, it's helpful to identify symmetry and the region it describes.

Step 2: Determine the Bounds for Integration

The region described is the circle's upper half since \(\sin \theta\) is positive in the first and second quadrants. As \(\theta\) varies from \(0\) to \(\pi\), \(r\) traces this semicircle.

Step 3: Set Up the Integral for Area

The formula for the area in polar coordinates is \(A = \frac{1}{2} \int_{a}^{b} r^2 \, d\theta\). Here, \(a = 0\) and \(b = \pi\), and \(r = 6 \sin \theta\). Plug these into the formula to get: \(A = \frac{1}{2} \int_{0}^{\pi} (6 \sin \theta)^2 \, d\theta\).

Step 4: Evaluate the Integral

Inside the integral, simplify: \((6 \sin \theta)^2 = 36 \sin^2 \theta\). Thus, \(A = 18 \int_{0}^{\pi} \sin^2 \theta \, d\theta\). Use the identity \(\sin^2 \theta = \frac{1 - \cos 2\theta}{2}\) to rewrite the integral: \(A = 18 \int_{0}^{\pi} \frac{1 - \cos 2\theta}{2} \, d\theta\). This becomes \(A = 9 \int_{0}^{\pi} (1 - \cos 2\theta) \, d\theta\).

Step 5: Solve the Integral

Compute the integral: \(A = 9 \left[ \theta - \frac{1}{2} \sin 2\theta \right]_{0}^{\pi}\). Evaluate: \(\theta\) from \(0\) to \(\pi\) is \(\pi\), and \(\sin 2\theta\) from \(0\) to \(\pi\) evaluates to 0. Therefore: \(A = 9(\pi - 0) = 9\pi\).

Step 6: Conclusion

The area enclosed by the polar curve \(r = 6 \sin \theta\) is the area of the semicircle, calculated as \(9\pi\).

Key Concepts

Area Calculation

When working with polar coordinates, calculating the area of a region enclosed by a polar curve involves integrating the square of the radius function. The standard formula used is:

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

This formula results from considering infinitesimally small sectors of the polar region.
Each sector can be viewed as a narrow triangle, and the integration calculates the sum of all these triangular areas over the specified range of angles.
In the given problem, you apply this formula after identifying that the radius function \(r = 6 \sin \theta\) traces the path of a semicircle.
To correctly compute the area, you need to carefully determine the proper bounds for \(\theta\), ensuring they cover the region entirely.

Polar Equation

A polar equation relates the radial distance \(r\) from the origin to the angle \(\theta\) measured from the positive x-axis. The equation \(r = 6 \sin \theta\) represents a circle centered above the origin in the x-y plane.

• It’s crucial to understand this representation to determine the graph's shape and position.
• Converting between polar and Cartesian coordinates helps visualize the graph.

Recognizing the shape and size of the region allows for correct area computation or transformation of equation forms.
The symmetry properties inherent in \(\sin\) and \(\cos\) functions, and polar representation, help simplify understanding complex graphs.
Visualizing these curves involves imagining how \(r\) varies as \(\theta\) changes, showing how the curve develops in the polar plane.

Integration Bounds

Determining the correct bounds for integration in polar coordinates is crucial. For the problem given, bounds are determined by the properties of the trigonometric function involved.
The sine function, \(\sin \theta\), is positive in the first and second quadrants, guiding the bounds set from \(\theta = 0\) to \(\theta = \pi\).

• This range ensures you capture the upper hemisphere of the circle, which traces the semicircle as \(\theta\) sweeps from 0 to \(\pi\).
• Choosing bounds that respect the nature of the function and the desired region is vital for accurate calculations.

Errors in bounds can lead to only partially computing the area or even integrating the unintended region, off setting the answer.

Trigonometric Identities

Trigonometric identities are pivotal in simplifying complex integrals in polar coordinates. In our scenario, simplifying \(\sin^2 \theta\) is essential to perform the integral.

• Using the identity \(\sin^2 \theta = \frac{1 - \cos 2\theta}{2}\), you can transform the function into a more integrable form.
• This identity helps break down powers of sine or cosine functions, leading to expressions easier to integrate over specified bounds.

Such simplifications reduce computational difficulty, avoiding potential errors and making solving these integrals feasible.
Mastering these identities enables one to solve varied problems efficiently, particularly those involving trigonometric integrals.