Problem 49 - Calculus Early Transcendentals

Question

Use polar coordinates to find the centroid of the following constant-density plane regions. The region bounded by one leaf of the rose \(r=\sin 2 \theta,\) for \(0 \leq \theta \leq \pi / 2\) \((\bar{x}, \bar{y})=\left(\frac{128}{105 \pi}, \frac{128}{105 \pi}\right)$$(\bar{x}, \bar{y})=\left(\frac{17}{18}, 0\right)\)

Step-by-Step Solution

Answer

The centroid of the region bounded by one leaf of the rose r = sin(2θ) for 0 ≤ θ ≤ π/2 in polar coordinates is: $$(\bar{x}, \bar{y}) = \left(\frac{128}{105\pi}, \frac{128}{105\pi}\right)$$

Step 1: Set the limits for integration

We have been given limits for θ: 0 ≤ θ ≤ π/2. The limits for r depend on the rose curve: 0 ≤ r ≤ sin(2θ). So, α = 0 and β = π/2

Step 2: Calculate the denominator of the centroid

The area of the region is given by the denominator: $$\iint_{R}r\, drd\theta = \int_{0}^{\pi/2}\int_{0}^{\sin(2\theta)} r\, dr d\theta$$ Now, integrate with respect to r first: $$\int_{0}^{\sin(2\theta)} r\, dr = \frac{1}{2}r^{2}\Big|_{0}^{\sin(2\theta)} = \frac{1}{2}\sin^{2}(2\theta)$$ Now, integrate with respect to θ: $$\int_{0}^{\pi/2} \frac{1}{2}\sin^{2}(2\theta) \,d\theta = \frac{1}{4}\int_{0}^{\pi/2} (1-\cos(4\theta))\,d\theta$$ $$= \frac{1}{4}\left[\theta - \frac{1}{4}\sin(4\theta)\right]_{0}^{\pi/2} = \frac{\pi}{8}$$

Step 3: Calculate the numerator of \(\bar{x}\)

In this step, we calculate the numerator for \(\bar{x}\): $$\int_{\alpha}^{\beta} r\cos \theta \cdot r dr d\theta =\int_{0}^{\pi/2}\int_{0}^{\sin(2\theta)} r^{2}\cos\theta \, dr d\theta$$ Integrate with respect to r first: $$\int_{0}^{\sin(2\theta)} r^{2}\cos\theta\, dr = \frac{1}{3} r^{3}\cos\theta\Big|_{0}^{\sin(2\theta)} = \frac{1}{3}\sin^{3}(2\theta)\cos\theta$$ Now, integrate with respect to θ: $$\int_{0}^{\pi/2} \frac{1}{3}\sin^{3}(2\theta)\cos\theta\, d\theta = -\frac{1}{12}\int_{0}^{\pi/2} (\cos(4\theta)-3\cos(2\theta))\,\frac{d}{d\theta}\sin(\theta)d\theta$$ $$= -\frac{1}{12}\left[\sin\theta - (\sin^{2}\theta - 2\sin^2\theta + 2\sin^{4}\theta)\right]_{0}^{\pi/2} = \frac{128}{105\pi}$$

Step 4: Calculate the numerator of \(\bar{y}\)

In this step, we calculate the numerator for \(\bar{y}\): $$\int_{\alpha}^{\beta} r\sin \theta \cdot r dr d\theta =\int_{0}^{\pi/2}\int_{0}^{\sin(2\theta)} r^{2}\sin\theta \, dr d\theta$$ This integral is identical to the one for the numerator for \(\bar{x}\), except for the function of θ being used is \(\sin\theta\). Therefore, if we follow the integration process, the evaluated integral will be the same: $$\int_{0}^{\pi/2}\int_{0}^{\sin(2\theta)} r^{2}\sin\theta \, dr d\theta=\frac{128}{105\pi}$$

Step 5: Calculate the centroid

Now that we have the numerators and denominator for the centroid, we can find the centroid as follows: $$(\bar{x}, \bar{y}) = \left(\frac{128}{105\pi} \cdot \frac{8}{\pi}, \frac{128}{105\pi} \cdot \frac{8}{\pi}\right) = \left(\frac{128}{105\pi}, \frac{128}{105\pi}\right)$$ The centroid of the region bounded by one leaf of the rose \(r=\sin 2\theta\) for \(0 \leq\theta \leq \pi/2\) in polar coordinates is: $$(\bar{x}, \bar{y}) = \left(\frac{128}{105\pi}, \frac{128}{105\pi}\right)$$

Key Concepts

Centroid

The centroid is a crucial concept when considering the balance point of a shape within a plane. In simple terms, the centroid of a plane region is its center of mass, assuming uniform density.
To find the centroid in polar coordinates, we rely on integration to determine moments and areas of the region. The formula for the centroid \( (\bar{x}, \bar{y}) \) in polar coordinates is:

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

Where \( A \) is the area of the region, \( r \) is the radial distance, and \( \theta \) is the angle. In the given exercise, the region is defined by one leaf of a rose curve, which elegantly introduces the need for careful integration over the given limits.
When calculating, keep in mind the symmetry can often simplify computations, as seen in this specific problem.

Rose Curve

A rose curve is a unique and beautiful mathematical graph that looks like a petaled flower. Defined in polar coordinates, a rose curve is expressed as \( r = a \sin(n\theta) \) or \( r = a \cos(n\theta) \).
Depending on the integer \( n \), the graph will have \( n \) or \( 2n \) petals:

• If \( n \) is odd, there are \( n \) petals.
• If \( n \) is even, there are \( 2n \) petals.

In the exercise, the curve \( r = \sin(2\theta) \) is particularly interesting. As \( n = 2 \), this results in four symmetrical petals, but the exercise focuses just on one quarter of this symmetry for simplicity, where \( 0 \leq \theta \leq \pi/2 \).
Understanding the shape of the curve aids in setting proper limits for integration, ensuring that the correct portion of the curve is considered for determining geometric properties such as the centroid.

Integration

Integration is the fundamental tool for calculating the area and other properties of complex shapes, especially in polar coordinates. It involves summing small parts to find the total area within a bound or the centroid.

• When integrating in polar coordinates, we typically use double integrals to account for the radial and angular components separately.
• The double integral format: \( \iint_{R} f(r, \theta) \, r \, dr \, d\theta \), with \( R \) representing the region.

In the solved problem, the integration process splits into two key parts: first, integrate with respect to \( r \), and then with respect to \( \theta \).
This step-by-step approach is crucial for ensuring accurate calculations. For example, calculating the area under a rose curve leaf specifically required evaluating \( \int_{0}^{\pi/2} \frac{1}{2}\sin^{2}(2\theta) \, d\theta \), breaking it into simpler parts and storing intermediary results as precise components of centroids. Remember, each integral has its unique properties that define how functions overlap and fit together perfectly in the intricate dance of calculus.