Question

Use polar coordinates to find the centroid of the following constant-density plane regions. The region bounded by the limaçon \(r=2+\cos \theta\)

Short answer

In summary, the area of the region bounded by the limaçon \(r=2+\cos \theta\) is \(A=\frac{9\pi}{4}\). The centroid of the region is located at \((\frac{8}{9}, 0)\). This is determined by calculating the x-coordinate centroid through the integral formula and finding the y-coordinate centroid to be zero since the bounded region is symmetric about the x-axis.

Step-by-step solution

1.1- Setting up the integral

To calculate the area of the region bounded by the limaçon \(r=2+\cos \theta\), we'll set up the integral: $$A = \frac{1}{2}\int_{\alpha}^{\beta} r^2(\theta) d\theta$$ Here, \(A\) is the area, and the limits of integration \(\alpha\) and \(\beta\) correspond to the angles where the limaçon curve starts and ends. It's important to note that the limaçon curve starts and ends at the intersection points of the curve with the positive x-axis. Solving for these points: \(r(0)=2+\cos 0=2+1=3\) \(r(\pi)=2+\cos \pi=2-1=1\) So, the limits of integration are as follows: $$A=\frac{1}{2}\int_{0}^{\pi} (2+\cos\theta)^2 d\theta$$

1.2- Evaluating the integral

Now that we have our integral set up, let's evaluate it: $$A=\frac{1}{2}\int_{0}^{\pi} (4+4\cos\theta+\cos^2\theta)d\theta$$ Using the trigonometric identity \(cos^2\theta=\frac{1}{2}(1+\cos 2\theta)\), we get: $$A=\frac{1}{2}\int_{0}^{\pi} (4+4\cos\theta+\frac{1}{2}(1+\cos 2\theta))d\theta$$ Now, integrate each term separately and evaluate from \(0\) to \(\pi\): $$A=\frac{1}{2}[4\theta+4\sin\theta+\frac{1}{2}\theta+\frac{1}{4}\sin 2\theta]\Big|_0^\pi$$

1.3- Calculation of the total area

After evaluating the integral, it's time to calculate the total area of the region: $$A=\frac{1}{2}(4\pi+0+\frac{1}{2}\pi+0) = \frac{9\pi}{4}$$ #Phase 2: Compute the centroid of the region using polar coordinates#

2.1- Setting up the x-coordinate formula

To find the centroid's x-coordinate, we'll use the following formula: $$x_{center}=\frac{1}{A}\int_{0}^{\pi} \frac{1}{2}(2+\cos\theta)^3 \cos\theta d\theta$$

2.2- Setting up the y-coordinate formula

To find the centroid's y-coordinate, we'll use the following formula: $$y_{center}=\frac{1}{A}\int_{0}^{\pi} \frac{1}{2}(2+\cos\theta)^3 \sin\theta d\theta$$

2.3- Evaluating the x-coordinate centroid

Now evaluate the x-coordinate centroid: $$x_{center}=\frac{1}{\frac{9\pi}{4}}\int_{0}^{\pi} \frac{1}{2}(2+\cos\theta)^3 \cos\theta d\theta$$ .Simplify the expression and solve the integral: $$x_{center}=\frac{4}{9\pi}\int_{0}^{\pi} (2+\cos\theta)^3 \cos\theta d\theta$$ Integrating and evaluating the integral, we get: $$x_{center}=\frac{8}{9}$$

2.4- Evaluating the y-coordinate centroid

Now evaluate the y-coordinate centroid: $$y_{center}=\frac{1}{\frac{9\pi}{4}}\int_{0}^{\pi} \frac{1}{2}(2+\cos\theta)^3 \sin\theta d\theta$$ Similarly, we simplify the expression and solve the integral: $$y_{center}=\frac{4}{9\pi}\int_{0}^{\pi} (2+\cos\theta)^3 \sin\theta d\theta$$ However, after integrating, we find that the y-coordinate centroid is equal to 0 because the bounded region is symmetric about the x-axis. $$y_{center}=0$$

2.5- Presenting the centroid

Finally, combining the x and y-coordinate centroids, we get the centroid of the region bounded by the limaçon \(r=2+\cos \theta\): Centroid: $$(\frac{8}{9}, 0)$$

Key concepts

Centroid Calculation

The centroid is essentially the "average" location of all the points in a shape. For our problem, we are looking at a limaçon curve in polar coordinates. Polar coordinates are a way of describing points on a plane using an angle and a distance from a central point. In this context, calculating the centroid involves finding its x and y coordinates using specific integral formulas.

Here is how it's done for polar coordinates:

• The x-coordinate of the centroid can be calculated by integrating the weighted x-location of the curve over its area.
• Similarly, the y-coordinate involves an integral over the weighted y-location of the shape.

By setting up and solving these integrals, we get precise x and y positions. In our example, we found:

• For x-coordinate: \[x_{center} = \frac{1}{A} \int_{0}^{\pi} \frac{1}{2}(2+\cos\theta)^3 \cos\theta \, d\theta\]and after solving, it equaled \(\frac{8}{9}\).
• For y-coordinate:\[y_{center} = \frac{1}{A} \int_{0}^{\pi} \frac{1}{2}(2+\cos\theta)^3 \sin\theta \, d\theta\]resulted in 0 due to symmetry in the curve.

Ultimately, the centroid for the region bounded by the limaçon curve \(r=2+\cos \theta\) is \(\left(\frac{8}{9}, 0\right)\). Calculating centroids is essential for understanding the balance and equilibrium within geometric shapes.

Integral Evaluation

Integral evaluation is crucial when working with continuous shapes and areas in calculus. To calculate an area under a curve, especially in polar coordinates, you integrate the function describing the curve over a specific range of angles.

For the shape we're examining (a limaçon), the area integral is set up as:\[A = \frac{1}{2}\int_{0}^{\pi} (2+\cos\theta)^2 \, d\theta\]To evaluate an integral like this:

• Start by expanding the squared term: \((2+\cos\theta)^2\) becomes \(4+4\cos\theta+\cos^2\theta\).
• Use trigonometric identities to simplify further. Here, \(\cos^2\theta\) is rewritten using the identity \(\cos^2\theta=\frac{1}{2}(1+\cos 2\theta)\).
• Once the expression is simplified, integrate each term separately.
• Finally, evaluate the definite integral using the limits, in our case, from 0 to \(\pi\).

Once evaluated, this integral provides the total area, which is essential not just for centroid calculations, but also for understanding the total "space" occupied by the limaçon curve. In our example, this area was found to be \(\frac{9\pi}{4}\). Solution of such integrals forms a powerful tool in solving problems of geometry and physics involving curves.

Limaçon Curve

A limaçon curve, in polar coordinates, is defined by equations of the form \(r = a + b \, \cos\theta\) or \(r = a + b \, \sin\theta\). These curves are fascinating because of their distinct, looped or dimpled shapes depending on the values of \(a\) and \(b\).

For our problem:

• The curve is given by \(r = 2 + \cos\theta\).
• This places it in the category of a limaçon with an inner loop because the coefficient in front of \(\cos\theta\) is smaller than the constant term, \(a = 2\) and \(b = 1\).
• The curve intersects itself at \(\theta =}\pi\), creating an inner loop.

These intersections define the shape's boundary. Understanding the properties of a limaçon is important because it tells us how the curve behaves, which affects calculations of areas and centroids. Limaçon curves can be applied in various advanced fields, such as antenna design and physics, due to their unique properties.

Symmetry in Geometry

Symmetry plays a pivotal role in simplifying geometric calculations. Symmetry refers to a balanced and proportionate similarity found in two halves of an object, that is, one half is the mirror image of the other half.

In the case of our limaçon curve \(r = 2 + \cos\theta\), the curve is symmetric about the x-axis:

• This means that any calculation involving vertical position (like the y-coordinate of the centroid) simplifies greatly.
• If a shape is symmetric about an axis, functions odd with respect to the axis will integrate to zero over symmetric limits. Hence, \(y_{center} = 0\) in our case.
• Such symmetry can drastically reduce computation time and complexity by reducing the number of calculations needed.

Realizing and applying symmetry in problems can both simplify and enhance the understanding of the geometry involved. Such concepts are significant in fields such as mechanical engineering and design, where exploiting symmetry can increase efficiency or performance of designs and structures.