Problem 71 - Essential Calculus

Question

For the region bounded by the graphs of the equations, find: (a) the volume of the solid formed by revolving the region about the \(x\) -axis and (b) the centroid of the region. $$ y=\sin x, y=0, x=0, x=\pi $$

Step-by-Step Solution

Answer

The volume of the solid formed by revolving the region around the x-axis is calculated first, followed by the coordinates of the centroid of the area. By evaluating the respective integrals, numerical values for these parameters can be obtained.

Step 1: Set up the integral for the volume

The boundaries for the region are \( x=0 \) and \( x=\pi \). The function describing the region is \( y = \sin(x) \). When revolving around the x-axis, the volume can be found using the formula: \[ V = \pi \int_a^b [y(x)]^2 dx \]where \( a = 0 \), \( b = \pi \) and \( y(x) = \sin(x) \) . So the volume integral becomes:\[ V = \pi \int_0^\pi [\sin(x)]^2 dx \]

Step 2: Calculate the volume

To solve the integral, apply a trigonometric identity for \( \sin^2(x) = \frac{1-\cos(2x)}{2} \). The volume integral becomes:\[ V = \pi \int_0^\pi \frac{1-\cos(2x)}{2} dx \]Solve the integral to find the volume.

Step 3: Set up the integrals for the centroid

The centroid (or center of mass) of a region divided by its area is defined by the coordinates \( (\bar{x},\bar{y}) \), where\[ \bar{x}=\frac{1}{A}\int_a^b x f(x) dx \]and \[ \bar{y}=\frac{1}{2A}\int_a^b f(x)^2 dx \]In these equations, \( A \) represents the area of the region and \( f(x) \) is the function describing the region. For our problem, the area A is given by\[ A = \int_0^\pi \sin(x) dx \]

Step 4: Calculate the centroid

To calculate the coordinates of the centroid, we substitute our known parameters into the earlier formulas and evaluate. We solve the integrals to derive the \( \bar{x} \) and \( \bar{y} \) values thereby giving the centroid of the region.

Key Concepts

Integral Calculus

Integral calculus is a fundamental part of calculus focusing on the accumulation of quantities, such as areas under curves and the total change over intervals. In this context, we use integral calculus to calculate the volume of a solid of revolution. This process involves an integral formula that revolves a region around an axis to create a 3-dimensional shape.
To find the volume in the exercise, we use the formula:

• \[ V = \pi \int_a^b [y(x)]^2 dx \]

Here, \(y(x)\) is the function being squared, which is \(\sin(x)\) in our example. We then integrate across the given boundaries from \(x=0\) to \(x=\pi\). This method helps us calculate the space inside the solid shape formed when the sine curve is rotated around the x-axis. Moral of the story: when dealing with volumes, always remember the powers of integral calculus!

Centroid Calculation

The centroid of a geometric shape is like its balance point, where the shape can be perfectly balanced in a uniform gravitational field. It can be determined using integrals, serving as the 'average position' for the shape's area.
In our task, the centroid coordinates \((\bar{x}, \bar{y})\) are calculated using:

• \[ \bar{x} = \frac{1}{A} \int_a^b x f(x) dx \]
• \[ \bar{y} = \frac{1}{2A} \int_a^b f(x)^2 dx \]

Here, \(A\) is the area of the region. For our sine boundary, \(A\) is calculated as \(\int_0^\pi \sin(x) dx\). These integrals help us find the exact point of balance for the area under the sine wave between \(x=0\) and \(x=\pi\). Calculating the centroid gives us insight into the distribution of the region's area, which can be crucial for understanding properties of physical bodies in engineering and physics.

Trigonometric Integration

Trigonometric integration involves integrating expressions containing trigonometric functions such as sine and cosine. It's essential when dealing with problems that involve periodic functions, like the sine function in our problem.
For instance, calculating the volume requires simplifying the integral of \( \sin^2(x) \) using a trigonometric identity:

• \[ \sin^2(x) = \frac{1 - \cos(2x)}{2} \]

This identity allows us to convert the original integral to a form that's easier to handle:

• \[ V = \pi \int_0^\pi \frac{1 - \cos(2x)}{2} dx \]

Breaking down trigonometric functions into simpler forms often simplifies integrals, making complex problems more approachable. Mastery of trigonometric integration is key to tackling various mathematical and engineering applications.