Question

Find the area of the described region. Enclosed by the circle: \(r=4 \sin \theta\)

Short answer

The area enclosed by the circle is \(8\pi\).

Step-by-step solution

Understand the Problem

We are asked to find the area enclosed by the circle given in polar coordinates as \( r = 4 \sin \theta \). This circle is centered at \( (0,2) \) with a radius of 2, representing a circle above the horizontal axis in terms of \( r \) and \( \theta \).

Identify the Area Formula in Polar Coordinates

The formula for the area \( A \) of a region in polar coordinates is given by: \[ A = \frac{1}{2} \int_{0}^{2\pi} r^2 d\theta \] where \( r \) is a function of \( \theta \).

Substitute the Function into the Area Formula

Substitute \( r = 4 \sin \theta \) into the area formula: \[ A = \frac{1}{2} \int_{0}^{2\pi} (4 \sin \theta)^2 d\theta \].

Simplify the Integral Expression

Simplify the expression: \[ A = \frac{1}{2} \int_{0}^{2\pi} 16 \sin^2 \theta \, d\theta \] \[ A = 8 \int_{0}^{2\pi} \sin^2 \theta \, d\theta \].

Use Trigonometric Identity to Simplify Further

Recall that \( \sin^2 \theta = \frac{1}{2} - \frac{1}{2}\cos(2\theta) \). Substitute this identity into the integral: \[ A = 8 \int_{0}^{2\pi} \left( \frac{1}{2} - \frac{1}{2} \cos(2\theta) \right) \, d\theta \] \[ = 8 \left[ \int_{0}^{2\pi} \frac{1}{2} \, d\theta - \int_{0}^{2\pi} \frac{1}{2} \cos(2\theta) \, d\theta \right] \].

Solve the Simplified Integrals

Calculate the integrals separately: For \( \int_{0}^{2\pi} \frac{1}{2} \, d\theta \), the result is \( \pi \). For \( \int_{0}^{2\pi} \frac{1}{2} \cos(2\theta) \, d\theta \), the result is \( 0 \) because the integral of a full wave of cosine over its period is zero.

Calculate the Final Area

Substitute back the evaluated integrals: \[ A = 8 \left[ \pi - 0 \right] = 8\pi \]. This represents the area enclosed by the given curve.

Key concepts

Integral in Polar Coordinates

When calculating the area of a region enclosed by a curve in polar coordinates, we can utilize the specific integral formula designed for this purpose. In polar coordinates, the position of a point is described by the radius (\( r \)) and angle (\( \theta \)) from the origin.

The area \( A \) of a region bounded by \( r = f(\theta) \), from \( \theta = \alpha \) to \( \theta = \beta \), can be computed using the integral:
\[A = \frac{1}{2} \int_{\alpha}^{\beta} [f(\theta)]^2 \, d\theta\]

This is derived from the geometry inherent in polar coordinates, where the segment of the circle swept by a small angle \( d\theta \) creates a sector of area. Thus, this integral sums up the areas of these infinite tiny sectors throughout the desired angular range.

• Ensure \( \alpha \) and \( \beta \) correctly reflect the part of the curve you're interested in. Often, for full circles, it will be from \( 0 \) to \( 2\pi \).
• Double-check your function \( f(\theta) \) before squaring and integrating to avoid errors in area calculation.

Trigonometric Identity

Trigonometric identities are essential tools that simplify complex expressions involving trigonometric functions. In this problem, finding the integral of \( \sin^2 \theta \) takes advantage of one such identity.

The identity used here is:
\[\sin^2 \theta = \frac{1}{2}(1 - \cos(2\theta))\]

This transforms the squared sine function into a more convenient form for integration. The individual components of this identity, namely \( 1 \) and \( \cos(2\theta) \), are straightforward to integrate over a full period \( 0 \) to \( 2\pi \), which is often required in polar coordinates calculations.

Benefits of this identity include:

• Simplification: Transforms a complex function into simpler integrands.
• Easy Integration: The resulting components are easy to integrate.

Without using such identities, the integration process would be far more cumbersome, leading to potential errors and increased effort.

Circle in Polar Coordinates

The representation of circles in polar coordinates is both elegant and intuitive when compared to Cartesian coordinates. A circle centered at the origin with radius \( a \) is simply given by \( r = a \). However, the equation \( r = 4 \sin \theta \) explored in the given problem represents a different scenario.

This equation describes a circle that is centered at \( (0,2) \) in Cartesian terms, with a radius of 2. Such a depiction arises by transforming polar representation into Cartesian coordinates. The conversion relationships are:
\[ x = r \cos \theta, \quad y = r \sin \theta \]

Upon substituting \( r = 4 \sin \theta \) into these equations, the circle's properties become evident:
- The total range for \( \theta \), often \( 0 \) to \( 2\pi \), helps to capture the full extent of the circle.
- The circle's placement is above the horizontal axis, confirming it sits centered vertically at \( y=2 \) due to the inclusion of the \( \sin \theta \) term.

• This understanding assists in correctly setting up the bounds and functions when plotting or calculating areas involving circles in polar contexts.

Overall, polar coordinates offer a powerful way to handle circular shapes more naturally and beautifully.