Problem 81 - Calculus: One and Several Variables

Question

Use a graphing utility to draw the curve \(y=1-\sin x\) from \(x=0\) to \(x=\pi .\) Then use a CAS (a) to find the area of the region \(\Omega\) between the curve ard the \(x\) -axis. (b) to find the volume of the solid generated by revolving \(\Omega\) about the \(y\) -axis. (c) to find the centroid of \(\Omega\)

Step-by-Step Solution

Answer

(a) The area of region Ω is approximately \(2\pi\). (b) The volume of the solid generated by revolving Ω about the y-axis is approximately \(\frac{5}{2} \pi^2\). (c) The centroid of region Ω is approximately located at \((\frac{\pi}{2}, \frac{2}{3})\).

Step 1: Graph the curve \(y = 1 - \sin x\) from \(x = 0\) to \(x = \pi\)

Using a graphing utility, plot the curve \(y = 1 - \sin x\) for the given domain \(x = 0\) to \(x = \pi\). You should observe a wave-like curve above the x-axis, with endpoints at \((0,1)\) and \((\pi, 0)\).

Step 2: Find the area of region \(\Omega\)

To find the area of region \(\Omega\), we will first set up the integral representing the area under the curve \(y = 1 - \sin x\) from \(x = 0\) to \(x = \pi\): \[A = \int_{0}^{\pi} (1 - \sin x) dx\] Now, use a Computer Algebra System (CAS) to evaluate the integral. You will find: \[A \approx 2\pi\]

Step 3: Find the volume of the solid generated by revolving \(\Omega\) about the y-axis

To find the volume of the solid generated by revolving \(\Omega\) about the y-axis, we will use the washer method. This involves the integral: \[V = \pi \int_{0}^{\pi} x^2 (1 - \sin x) dx\] Again, use a Computer Algebra System (CAS) to evaluate the integral. You will find: \[V \approx \frac{5}{2} \pi^2\]

Step 4: Find the centroid of region \(\Omega\)

The centroid of region \(\Omega\) can be found using the following formulas for the x and y coordinates of the centroid, which involve the integral calculations for the area of region \(\Omega\): \[ \bar{x} = \frac{1}{A} \int_{0}^{\pi} x (1 - \sin x) dx \] \[ \bar{y} = \frac{1}{A} \int_{0}^{\pi} \frac{1}{2} (1 - \sin x)^2 dx \] Using a Computer Algebra System (CAS) to evaluate these integrals, and plugging in the area found above, you will find: \[\bar{x} \approx \frac{\pi}{2}\] \[\bar{y} \approx \frac{2}{3}\] So the centroid of region \(\Omega\) is approximately located at \((\frac{\pi}{2}, \frac{2}{3})\).

Key Concepts

Area Under Curve

The area under a curve in the context of integral calculus represents the total "accumulated" space between the curve and the horizontal axis within specified bounds. In this case, we are looking at the region between the curve defined by \(y = 1 - \sin x\) and the x-axis from \(x = 0\) to \(x = \pi\). To find this area, we use the definite integral, which calculates this "accumulated" quantity.

The formula for finding the area under this curve is given by:

• \[ A = \int_{0}^{\pi} (1 - \sin x) \ dx \]

This integral essentially sums up all the tiny slices of the space under the curve from the starting point \(x = 0\) to the endpoint \(x = \pi\).

By evaluating this integral using a tool like a Computer Algebra System (CAS), we find that the area \(A\) is approximately \(2\pi\). This represents the "size" of the space below the curve and above the x-axis within the indicated interval.

Understanding this concept of area under a curve is fundamental because it can be used to measure various real-world phenomena such as distance traveled over time, total growth over an interval, or any quantity that is cumulative in nature.

Volume of Revolution

The volume of revolution is a method in integral calculus for finding the volume of a 3-dimensional object created by rotating a 2-dimensional area about an axis. To solve this, the washer method is often employed when we revolve around vertical or horizontal lines not touching the curve.

Here, the region \(\Omega\) under the curve \(y = 1 - \sin x\) from \(x = 0\) to \(x = \pi\) is revolved around the y-axis to form a solid of revolution. The washer method involves calculating the integral:

• \[ V = \pi \int_{0}^{\pi} x^2 (1 - \sin x) \ dx \]

This integral calculates the volume by "stacking" washers (disks with holes) of infinitesimal thickness and summing their volumes from \(x = 0\) to \(x = \pi\).

After evaluating the integral with a CAS, it is found that the volume \(V\) is approximately \(\frac{5}{2} \pi^2\). This calculated volume represents the total space inside the 3-dimensional structure obtained by rotating the region \(\Omega\) about the y-axis.

Centroid of a Region

The centroid of a region in integral calculus is the point that serves as the "average" position of the region. It is essentially the balancing point of a planar region. For the region \(\Omega\), defined by the curve \(y = 1 - \sin x\) from \(x = 0\) to \(x = \pi\), the centroid can be located using two main formulas for the x and y coordinates:

• The x-coordinate of the centroid, \(\bar{x}\), is given by:\[ \bar{x} = \frac{1}{A} \int_{0}^{\pi} x (1 - \sin x) \ dx \]
• The y-coordinate of the centroid, \(\bar{y}\), uses:\[ \bar{y} = \frac{1}{A} \int_{0}^{\pi} \frac{1}{2} (1 - \sin x)^2 \ dx \]

These integrals essentially provide a weighted average of the x and y positions of the region, with the area \(A\) previously calculated in the problem.

After evaluating these integrals with a CAS, the centroid is approximately found at \((\frac{\pi}{2}, \frac{2}{3})\). This means that if you were to balance the region on a point, it would balance perfectly at this location, indicating the region's center of mass.