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. $$ y=\sin x, y=0, x=0, x=\pi $$

Short 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-by-step solution

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:\[ V = \pi \int_0^\pi [\sin(x)]^2 dx \]

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.

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\[ A = \int_0^\pi \sin(x) dx \]

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.