Question

Use integration to show that the circles \(r=2 a \cos \theta\) and \(r=2 a \sin \theta\) have the same area, which is \(\pi a^{2}\)

Short answer

" Yes, the circles with equations \(r = 2a \cos \theta\) and \(r = 2a \sin \theta\) have the same area. The area of both circles is \(\pi a^2\).

Step-by-step solution

Find the limits of integration for both equations

For both circle equations, we can see that they are defined for \(0 \leq \theta \leq \pi\) to cover the entire area enclosed by each circle.

Convert the polar equations into area integrals

We will use the polar area formula: \(A = \frac{1}{2} \int_\alpha^\beta r^2 d\theta\), where \(\alpha\) and \(\beta\) are the limits of integration. For the first circle \(r = 2a \cos \theta\), the area integral will be: \(A_1 = \frac{1}{2} \int_0^\pi (2a\cos \theta)^2 d\theta\) For the second circle \(r = 2a \sin \theta\), the area integral will be: \(A_2 = \frac{1}{2} \int_0^\pi (2a\sin \theta)^2 d\theta\)

Evaluate the area integrals

Let's evaluate the area integrals one by one: For the first circle, \(A_1 = \frac{1}{2} \int_0^\pi (4a^2\cos^2 \theta) d\theta\) Now we can use the identity \(\cos^2 \theta = \frac{1}{2}(1 + \cos 2\theta)\) to simplify the integral: \(A_1 = \frac{1}{2} \cdot 4a^2 \int_0^\pi \frac{1}{2}(1 + \cos 2\theta) d\theta\) \(A_1 = 2a^2 \int_0^\pi \frac{1}{2}(1 + \cos 2\theta) d\theta\) Now, we can integrate: \(A_1 = 2a^2 \left[\frac{1}{2}(\theta + \frac{1}{4}\sin 2\theta) \right]_0^\pi\) Evaluate the integral at the limits: \(A_1 = 2a^2 [\frac{1}{2}(\pi - 0 + 0)]\) \(A_1 = \pi a^2\) For the second circle, \(A_2 = \frac{1}{2} \int_0^\pi (4a^2\sin^2 \theta) d\theta\) We can use the identity \(\sin^2 \theta = \frac{1}{2}(1 - \cos 2\theta)\): \(A_2 = \frac{1}{2} \cdot 4a^2 \int_0^\pi \frac{1}{2}(1 - \cos 2\theta) d\theta\) \(A_2 = 2a^2 \int_0^\pi \frac{1}{2}(1 - \cos 2\theta) d\theta\) Now, we can integrate: \(A_2 = 2a^2 \left[\frac{1}{2}(\theta - \frac{1}{4}\sin 2\theta) \right]_0^\pi\) Evaluate the integral at the limits: \(A_2 = 2a^2 [\frac{1}{2}(\pi - 0 - 0)]\) \(A_2 = \pi a^2\) We can see that both \(A_1\) and \(A_2\) are equal to \(\pi a^2\). Therefore, the circles \(r=2a\cos \theta\) and \(r=2a\sin \theta\) have the same area, which is \(\pi a^{2}\).

Key concepts

Polar Coordinates

Polar coordinates provide an alternative way to describe points in the plane, distinct from the Cartesian coordinate system, which uses \(x, y\) coordinates. In polar coordinates, each point is determined by a radius and an angle with respect to a fixed reference direction, typically the positive x-axis.
Key components include:

• The radius \(r\): The distance from the origin to the point.
• The angle \(\theta\): Measured in radians, the angle from the positive x-axis to the line connecting the origin to the point.

Using polar coordinates is particularly advantageous for plotting curves and solving problems involving circular or radial symmetry. For instance, the equations \(r = 2a \cos \theta\) and \(r = 2a \sin \theta\) represent circles in the polar system. Here, the radius changes as the angle changes, forming a curve or shape such as a circle.

Area Under a Curve

The concept of finding the area under a curve is a fundamental idea in calculus, often solved with the concept of integration. When dealing with polar coordinates, the area under a curve is found using the polar area formula:
\[A = \frac{1}{2} \int_\alpha^\beta r^2 d\theta\]
This formula calculates the area enclosed by a polar curve between two angles, \(\alpha\) and \(\beta\). Here's why it works:

• The integral sums up infinitely small sectors of the circle.
• Each sector’s area is analogous to a slice of pie, with area \(\frac{1}{2} r^2 d\theta\).

Using the exercises provided, \(r = 2a \cos \theta\) and \(r = 2a \sin \theta\) were converted into integrals that determined their areas. Integrating trigonometric functions like \(\cos^2 \theta\) and \(\sin^2 \theta\), we proved the area enclosed is \(\pi a^2\).

Trigonometric Identities

Trigonometric identities are useful tools in simplifying expressions and solving equations involving trigonometric functions. In the context of this problem, they help to solve integrals that appear complex at first sight. The main identities used are:

• \( \cos^2 \theta = \frac{1}{2}(1 + \cos 2\theta) \)
• \( \sin^2 \theta = \frac{1}{2}(1 - \cos 2\theta) \)

These identities are derived from the double angle formulas and allow us to express the square of sine or cosine in terms of a linear form. This reformulation is important, as it turns complex trigonometric expressions into simpler ones that are easy to integrate.
For example, applying these identities allows the integrals \( \int (\cos^2 \theta) d\theta \) and \( \int (\sin^2 \theta) d\theta \) to transform into expressions that integrate to well-known results, ultimately simplifying the evaluation of the areas and confirming the circles have equal area.