Question

Use polar coordinates to find the centroid of the following constant-density plane regions. The semicircular disk \(R=\\{(r, \theta): 0 \leq r \leq 2,0 \leq \theta \leq \pi\\}\)

Short answer

Based on the step by step solution, the centroid of the given semicircular disk with a constant-density plane region is at the point (\(\frac{4}{3}\), \(\frac{\pi}{2}\)) in polar coordinates. This means that the centroid is located at an average distance of \(\frac{4}{3}\) from the origin and at an angle of \(\frac{\pi}{2}\) from the positive x-axis.

Step-by-step solution

Finding Area of the Region

Since the given region is a semicircular disk with radius 2, its area A can be calculated as: \[A = \frac{1}{2}\pi(2)^2 = 2\pi\]

Calculate \(\overline{r}\) and \(\overline{\theta}\)

We'll compute the average distance of all points from the origin, \(\overline{r}\), and the average angle made by these points with the positive x-axis, \(\overline{\theta}\): \[\overline{r} = \frac{1}{A}\int_0^{2}\int_0^{\pi} r \cdot r d\theta dr\] \[\overline{\theta} = \frac{1}{A}\int_0^{2}\int_0^{\pi} r \theta d\theta dr\]

Evaluate the integrals

First, we will evaluate the integral for \(\overline{r}\): \[\overline{r} = \frac{1}{2\pi}\int_0^{2}\int_0^{\pi} r^2 d\theta dr\] \[= \frac{1}{2\pi}\int_0^{2} r^2 \int_0^{\pi} d\theta dr\] \[= \frac{1}{2\pi}\int_0^{2} r^2 (\pi - 0) dr\] \[= \frac{1}{2}\int_0^{2} r^2 dr\] \[= \frac{1}{2} \left[\frac{1}{3}r^3\right]_0^2\] \[= \frac{1}{3}(2)^3 = \frac{4}{3}\] Now, we will evaluate the integral for \(\overline{\theta}\): \[\overline{\theta} = \frac{1}{2\pi}\int_0^{2}\int_0^{\pi} r \theta d\theta dr\] \[= \frac{1}{2\pi}\int_0^{2} r \left[\frac{1}{2}\theta^2\right]_0^\pi dr\] \[= \frac{1}{2\pi}\int_0^{2} r \left(\frac{1}{2}\pi^2\right) dr\] \[= \frac{\pi}{4}\int_0^{2} r dr\] \[= \frac{\pi}{4} \left[\frac{1}{2}r^2\right]_0^2\] \[= \frac{\pi}{4}(2)^2 = \frac{\pi}{2}\]

Compute Centroid

The centroid of the given semicircular disk is thus at the point (\(\overline{r}\), \(\overline{\theta}\)): \[(\frac{4}{3}, \frac{\pi}{2})\] In polar coordinates, the centroid is located at the average distance \(\frac{4}{3}\) from the origin and at an angle \(\frac{\pi}{2}\) from the positive x-axis.

Key concepts

Centroid Calculation

When determining the centroid of a shape, it is all about finding the "center of mass" or the balance point. For shapes with uniform density, like a constant-density region, this job becomes straightforward. The centroid is a vital concept in geometry and physics. It helps in understanding the distribution of mass and designing structures.

• The centroid is analogous to the center point in a shape.
• For symmetrical shapes, the centroid lies along the axis of symmetry.

To find the centroid in polar coordinates, which is sometimes more descriptive than Cartesian coordinates, we use integrals to average the position's radii and angles. Here, for a semicircular disk, it will expand our understanding beyond linear coordinates.
The exercise involves calculating \(\overline{r}\), the average radial distance, and \(\overline{\theta}\), the average angle.

Semicircular Disk

A semicircular disk is half of a full circle. Think of it as slicing a pizza into half and considering just the top or bottom part.

• The boundary of a semicircular disk forms a curve, or an arc, lying flat along a straight edge - the diameter.
• Its symmetry around the diameter simplifies many calculations.

In the given problem, the semicircular disk has a radius of 2. This means its entire geometry is derived from this radius. The area covered by the disk is calculated using the formula for the area of a circle but only for half of it. Thus, \(A = \frac{1}{2}\pi(2)^2 = 2\pi\). Understanding shapes like a semicircular disk is crucial for applying geometrical concepts to physical objects.

Integrals

Integration serves as a powerful tool in calculus for finding areas, volumes, and centroids. When working with polar coordinates, the concept might seem challenging at first because it involves radial lines originating from a point rather than rectangular grids.
Integrals allow us to accumulate small elements across a defined region under consideration.

• For \(\overline{r}\), we used \(\int_0^{2}\int_0^{\pi} r^2 d\theta dr\)
• For \(\overline{\theta}\), the setup was \(\int_0^{2}\int_0^{\pi} r \theta d\theta dr\)

Each integral has specific limits defined by the shape's characteristics. The inner integral first handles the angle, \(\theta\), followed by the radial \(r\). When solved step-by-step, like in the provided solution, they become more manageable and reveal the average position values essential for finding the centroid.

Constant-Density Region

In physics and geometry, a region with constant density is one where the distribution of mass is uniform. In simpler terms, each part of the area has the same weight per unit.
This assumption supports the simplicity of calculations for centroids because variations in density do not need to be considered. It's like imagining a perfectly balanced seesaw due to equal weight distribution over its span.

• This property makes it easier to use formulas and integral expressions without adjusting for mass differences.
• The focus remains solely on the shape and geometry of the region itself.

For the semicircular disk exercise, dealing with constant density means it's enough to consider radial and angular contributions to find the centroid without extra complications for changes in mass across the area.