Question

Use polar coordinates to find the centroid of the following constant-density plane regions. The region bounded by the cardioid \(r=1+\cos \theta\)

Short answer

Based on the solution provided, the centroid of the region bounded by the cardioid \(r=1+\cos\theta\) is at the coordinates \((0, \frac{4}{3})\).

Step-by-step solution

1. Find the area of the region

To find the area of the region bounded by the cardioid \(r=1+\cos\theta\), we'll use the polar area formula: \(A = \frac{1}{2}\int_{\alpha}^{\beta} r^2 d\theta\) Here, \(\alpha\) and \(\beta\) are the limits of integration in terms of \(\theta\). As the cardioid is fully contained in the range \(0 \leq \theta \leq 2\pi\), our limits of integration will be \(\alpha=0\) and \(\beta=2\pi\). So, \(A = \frac{1}{2}\int_{0}^{2\pi} (1+\cos\theta)^2 d\theta\)

2. Evaluate the area integral

First, let's expand the integrand: \((1+\cos\theta)^2 = 1 + 2\cos\theta + \cos^2\theta\) Now let's compute the integral: \(A = \frac{1}{2}\int_{0}^{2\pi} (1+2\cos\theta+\cos^2\theta) d\theta\) We can break this integral into three parts and calculate them separately: \(A = \frac{1}{2}\left(\int_{0}^{2\pi} d\theta + 2\int_{0}^{2\pi} \cos\theta d\theta + \int_{0}^{2\pi} \cos^2\theta d\theta\right)\) The first integral is straightforward: \(\int_{0}^{2\pi} d\theta = 2\pi\) The second integral involves \(\cos\theta\): \(\int_{0}^{2\pi} \cos\theta d\theta = \sin\theta |_0^{2\pi} = 0\) The third integral involves \(\cos^2\theta\). We will use the double-angle formula to rewrite it as: \(\frac{1}{2}(1 + \cos(2\theta))\) So, \(\int_{0}^{2\pi} \cos^2\theta d\theta = \frac{1}{2}\int_{0}^{2\pi} (1+\cos(2\theta)) d\theta\) After calculating the last integral, we have: \(\int_{0}^{2\pi} \cos^2\theta d\theta = \frac{1}{2}(\int_{0}^{2\pi} d\theta + \int_{0}^{2\pi} \cos(2\theta) d\theta) = \frac{1}{2}(2\pi) = \pi\) Therefore, \(A = \frac{1}{2}(2\pi + 0 + \pi) = \frac{3}{2}\pi\)

3. Find the centroid coordinates

Next, we need to find the centroid coordinates. For polar coordinates, these are given by the following formulas: \(\bar{x} = \frac{1}{A}\int_{\alpha}^{\beta} x r d\theta = \frac{1}{A}\int_{\alpha}^{\beta} (r\cos\theta)r d\theta\) \(\bar{y} = \frac{1}{A}\int_{\alpha}^{\beta} y r d\theta = \frac{1}{A}\int_{\alpha}^{\beta} (r\sin\theta)r d\theta\) Plugging in the values for \(\alpha\), \(\beta\), \(A\), and \(r\): \(\bar{x} = \frac{2}{3\pi}\int_{0}^{2\pi} (1+\cos\theta)(1+\cos\theta)\cos\theta d\theta\) \(\bar{y} = \frac{2}{3\pi}\int_{0}^{2\pi} (1+\cos\theta)(1+\cos\theta)\sin\theta d\theta\)

4. Evaluate the centroid coordinates integrals

Now we must evaluate these integrals: \(\bar{x} = \frac{2}{3\pi}\int_{0}^{2\pi} (1+\cos\theta)(1+\cos\theta)\cos\theta d\theta = \frac{2}{3\pi}\int_{0}^{2\pi} (1+2\cos\theta+\cos^2\theta)\cos\theta d\theta\) \(\bar{y} = \frac{2}{3\pi}\int_{0}^{2\pi} (1+\cos\theta)(1+\cos\theta)\sin\theta d\theta = \frac{2}{3\pi}\int_{0}^{2\pi} (1+2\cos\theta+\cos^2\theta)\sin\theta d\theta\) By calculating these integrals, we obtain: \(\bar{x} = \frac{2}{3\pi}\int_{0}^{2\pi} (\cos\theta+2\cos^2\theta+\cos^3\theta) d\theta = 0\) \(\bar{y} = \frac{2}{3\pi}\int_{0}^{2\pi} (\sin\theta+2\cos\theta\sin\theta+\cos^2\theta\sin\theta) d\theta = \frac{4}{3}\) So the centroid of the region bounded by the cardioid \(r=1+\cos\theta\) is at coordinates \((0, \frac{4}{3})\).

Key concepts

Centroid Calculation

A centroid is a crucial concept in geometry, helping us find the center of mass of a plane region with uniform density. For a region defined in polar coordinates, the centroid's coordinates \((\bar{x}, \bar{y})\) are derived through integration.
To find the centroid, we'll need the area \(A\) and the weighted coordinates from which the centroid formula is calculated:

• \(\bar{x} = \frac{1}{A}\int_{0}^{2\pi} (r \cos\theta) r \, d\theta\)
• \(\bar{y} = \frac{1}{A}\int_{0}^{2\pi} (r \sin\theta) r \, d\theta\)

In this context, \(r\) represents the radial distance from the pole. By evaluating these integrals, we find the coordinates \((0, \frac{4}{3})\). This means that the center of mass for the bounded region lies on the y-axis, above the origin.

Cardioid Equation

The cardioid equation \(r = 1 + \cos\theta\) represents a heart-shaped curve in polar coordinates. With its distinctive shape, the cardioid is symmetric around the polar axis (the x-axis in Cartesian coordinates).
Key features of the cardioid include:

• The distance from the origin varies as the angle \(\theta\) changes.
• At \(\theta = 0, 2\pi\), the maximum distance \(r=2\).
• At \(\theta = \pi\), \(r=0\), marking the cusp of the curve.

Understanding this equation helps in visualizing the region bounded by the cardioid, which is essential for further calculations involving area and centroid location.

Area of Polar Region

To find the area of a region bounded by a polar curve, we use the formula: \(A = \frac{1}{2}\int_{\alpha}^{\beta} r^2 d\theta\).
For the cardioid \(r = 1 + \cos\theta\), the integration limits are \(0\) to \(2\pi\).
The calculation involves:

• Squaring the function: \(r^2 = (1 + \cos\theta)^2\).
• Expanding and integrating: \((1 + 2\cos\theta + \cos^2\theta)\).
• Using trigonometric identities, such as \(\cos^2\theta = \frac{1}{2}(1 + \cos 2\theta)\).
• Evaluating each part separately to obtain \(\frac{3}{2}\pi\).

This calculated area is crucial for determining subsequent properties like the centroid.

Integration in Polar Coordinates

Integration in polar coordinates is essential for calculating properties like area and centroid. By converting Cartesian integrals into polar form, we can efficiently handle circular and heart-shaped regions.
The process involves:

• Setting up the integral with \(r^2\) to find area or \((r \cos\theta) r\) and \((r \sin\theta) r\) for centroids.
• Determining the limits of \(\theta\), which define the bounded region.
• Solving integrals using trigonometric identities like \(\cos^2\theta = \frac{1}{2}(1 + \cos 2\theta)\).

This technique simplifies complex shape calculations, allowing for elegant mathematical solutions. To illustrate, evaluating the area under a cardioid became straightforward using \(\frac{1}{2}\int_{0}^{2\pi} r^2 d\theta\), demonstrating the power of polar integration.