Question

Areas of circles 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

Answer: Yes, the areas of the circles are equal, as both have an area of \(\pi a^2\).

Step-by-step solution

Convert Polar Equations to Cartesian Coordinates

First, we need to convert the polar equations into Cartesian coordinates. The conversion formulas are: \(x = r\cos\theta\) \(y = r\sin\theta\) For circle 1 (\(r = 2a\cos\theta\)): \(x = (2a\cos\theta)\cos\theta = 2a\cos^2\theta\) \(y = (2a\cos\theta)\sin\theta = 2a\sin\theta\cos\theta\) For circle 2 (\(r = 2a\sin\theta\)): \(x = (2a\sin\theta)\cos\theta = 2a\sin\theta\cos\theta\) \(y = (2a\sin\theta)\sin\theta = 2a\sin^2\theta\)

Determine the Bounds of Integration

Next, we need to determine the bounds of integration for both circles. The polar equations represent circles centered at \((a, 0)\) and \((0, a)\) with radius \(a\). So, the limits of integration for \(\theta\) will be from \(0\) to \(\pi\) for both circles.

Set up the Integrals for Area

To find the area of each circle, we'll use the polar area formula: \(A = \frac{1}{2}\int_{\alpha}^{\beta} r^2 d\theta\) For circle 1 (\(r = 2a\cos\theta\)): \(A_1 = \frac{1}{2}\int_{0}^{\pi} (2a\cos\theta)^2 d\theta\) For circle 2 (\(r = 2a\sin\theta\)): \(A_2 = \frac{1}{2}\int_{0}^{\pi} (2a\sin\theta)^2 d\theta\)

Evaluate the Integrals

Now, we'll evaluate both integrals: For \(A_1\): \(A_1 = \frac{1}{2}\int_{0}^{\pi} (2a\cos\theta)^2 d\theta\) \(A_1 = 2a^2\int_{0}^{\pi} \cos^2\theta d\theta\) Using the identity \(\cos^2\theta = \frac{1}{2}(1 + \cos(2\theta))\): \(A_1 = 2a^2\int_{0}^{\pi} \frac{1}{2}(1 + \cos(2\theta)) d\theta\) \(A_1 = a^2\int_{0}^{\pi} (1 + \cos(2\theta)) d\theta\) \(A_1 = a^2\left[\theta + \frac{1}{2}\sin(2\theta)\right]_0^\pi\) \(A_1 = a^2\left[\pi + 0\right] = \pi a^2\) For \(A_2\): \(A_2 = \frac{1}{2}\int_{0}^{\pi} (2a\sin\theta)^2 d\theta\) \(A_2 = 2a^2\int_{0}^{\pi} \sin^2\theta d\theta\) Using the identity \(\sin^2\theta = \frac{1}{2}(1 - \cos(2\theta))\): \(A_2 = 2a^2\int_{0}^{\pi} \frac{1}{2}(1 - \cos(2\theta)) d\theta\) \(A_2 = a^2\int_{0}^{\pi} (1 - \cos(2\theta)) d\theta\) \(A_2 = a^2\left[\theta - \frac{1}{2}\sin(2\theta)\right]_0^\pi\) \(A_2 = a^2\left[\pi - 0\right] = \pi a^2\)

Compare the Areas of the Circles

As we see from our calculations, both circle 1 and circle 2 have the same area: \(A_1 = \pi a^2\) \(A_2 = \pi a^2\) Which is the desired result.

Key concepts

Polar Coordinates

Polar coordinates are a way to describe a point's position in a plane using a distance and an angle. Unlike the usual Cartesian coordinates that use two distances (x and y), polar coordinates are based on:

• a radius, \( r \), which is the distance from the origin (center of the circle) to the point, and
• an angle, \( \theta \), which is measured from the positive x-axis to the direction of the radius.

In polar coordinates, equations can often be simpler and more intuitive for circular shapes and rotations. In the original exercise, the circles were given as either \( r = 2a\cos\theta \) or \( r = 2a\sin\theta \). These equations define circles centered along the x-axis and y-axis, respectively. Understanding these representations can be very helpful in visualizing geometric patterns and simplifying the math involved in problems related to circles or other circular patterns.

Area of a Circle

Finding the area of a circle using calculus and integration can initially seem daunting, but it's more intuitive when broken down. The goal is to calculate how much space is inside the circle. The polar area formula is instrumental in this process:

\[ A = \frac{1}{2} \int_\alpha^\beta r^2 d\theta \]

For the exercise, the radii of the circles were \( r = 2a\cos\theta \) and \( r = 2a\sin\theta \). By squaring these radii and integrating between suitable bounds for \(\theta\) (in this case, from 0 to \(\pi\)), you can find the area. The factor of \( \frac{1}{2} \) accounts for the geometry of the circle. Calculating the integral yields the area \( \pi a^2 \), which is consistent with the well-known formula for the area of a circle with radius \( a \). This approach shows how calculus can generalize familiar concepts and provide a deeper understanding of geometry.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for any value of the variable where both sides of the equation are defined. These identities are incredibly useful in simplifying and solving integrals involving trigonometric terms. For the given exercise, two key identities were used:

• The identity for \( \cos^2\theta \): \[ \cos^2\theta = \frac{1}{2}(1 + \cos(2\theta))\] This helps when integrating \( \cos^2\theta \) over a region.
• The identity for \( \sin^2\theta \): \[ \sin^2\theta = \frac{1}{2}(1 - \cos(2\theta))\]This is useful for integrating \( \sin^2\theta \).

By converting \( \sin^2\theta \) and \( \cos^2\theta \) into simpler forms, these identities assist in breaking down complex integrals into manageable parts. Using these identities, one can integrate trigonometric functions more easily, ultimately showing how the areas of the circles become \( \pi a^2 \). These tools are essential in the field of calculus, especially when working with periodic functions typical in circular and wave-like phenomena.