Problem 70 - Essential Calculus

Question

Given the region bounded by the graphs of \(y=x \sin x, y=0, x=0,\) and \(x=\pi,\) find (a) the volume of the solid generated by revolving the region about the \(x\) -axis. (b) the volume of the solid generated by revolving the region about the \(y\) -axis. (c) the centroid of the region.

Step-by-Step Solution

Answer

The volume of the solid generated by revolving the region about the x-axis and the y-axis, as well as the coordinates of the centroid would be obtained by numerically solving the integrals set up in steps 1, 2 and 3 respectively. Due to the complexity of these integrals, a closed form solution is not simple to obtain and thus numerical solutions are preferred.

Step 1: Volume of solid around x-axis

First, calculate the volume of the solid generated by revolving the region about the x-axis. This uses the disc method. The generic formula is given by: \( V = \pi \int_{a}^{b} [f(x)]^2 dx \). Substituting \(f(x)=x \sin x\) gives: \( V_1 = \pi \int_{0}^{\pi} [x \sin x]^2 dx \). Solve this integral numerically to obtain the volume.

Step 2: Volume of solid around y-axis

Next, calculate the volume of the solid generated by revolving the region about the y-axis. This involves the cylindrical shell method. Solving for x in the equation \(y=x \sin x\) is nontrivial, thus a numerical method is preferred to solve this part. The volume is given as \(V_2 = 2 \pi \int_0^{b} [x \cdot f(x)] dx\). Substituting \(f(x)=x \sin x\) gives \(V_2 = 2 \pi \int_0^\pi [x \cdot x \sin x] dx\). Numerically solve this integral to obtain the volume.

Step 3: Centroid of the region

Finally, calculate the centroid of the region. The centroid coordinates are given by \((\overline{x}, \overline{y})\), where \(\overline{x} = \frac{1}{A} \cdot \int_{a}^{b} [x \cdot f(x)] dx\) and \(\overline{y} = \frac{1}{2A} \cdot \int_{a}^{b} [f(x)]^2 dx\). The area A in this case is: \(A = \int_0^\pi [x \sin x] dx\). These integrals yield the \(x\) and \(y\) coordinates of the centroid when computed numerically, where \(\overline{x}, \overline{y}\) are centroid on the x and y-axis, respectively.

Key Concepts

Disk Method

The Disk Method is a powerful technique in calculus used to find the volume of a solid of revolution. When you rotate a region around the x-axis, you imagine slicing the solid into thin disks. Each disk resembles a small circle with radius equal to the function value at that point.

To find the volume of the solid, you sum up the volumes of all these disks. The formula used is:

• \( V = \pi \int_{a}^{b} [f(x)]^2 dx \).

In this formula, \(f(x)\) is the radius of each disk, and you integrate from the starting point \(a\) to the ending point \(b\). This is exactly what you use when you have the region given by \(y = x \sin x\), \(y = 0\), from \(x = 0\) to \(x = \pi\).

By placing these values into the formula, you get an expression that can be solved to find the volume. When you substitute \(f(x) = x \sin x\), you compute \( \pi \int_{0}^{\pi} [x \sin x]^2 dx \). Numerical methods are often employed when the integration becomes complicated, as is the case here.

Shell Method

The Shell Method provides an alternative way to compute volumes of solids of revolution, especially useful when rotating around the y-axis. In this method, you imagine the solid as being made up of cylindrical shells instead of disks.

For each shell, you calculate the volume and add them up. The formula to find this volume is:

• \( V = 2\pi \int_{a}^{b} x \cdot f(x) dx \).

Here, \(x\) represents the radius of the shell and \(f(x)\) is the height of the shell. Integrating this product gives the volume of the solid.

Applying this to the problem, the function \(f(x) = x\sin x\) results in a complex integration that is typically solved numerically. You end up solving \( 2\pi \int_{0}^{\pi} [x \cdot x\sin x] dx \). Numerically solving the integral provides the volume of the solid formed by rotating the bounded region around the y-axis.

Centroid Calculation

The centroid, often referred to as the center of mass, of a region in a plane provides an average position of all points in that shape. In calculus, finding a centroid can involve some complex integration, but is crucial for understanding the balance point of something like a lamina or 2D shape.

To calculate the centroid, we use two formulas to find coordinates \( (\overline{x}, \overline{y}) \):

• \(\overline{x} = \frac{1}{A} \cdot \int_{a}^{b} x \cdot f(x) dx\)
• \(\overline{y} = \frac{1}{2A} \cdot \int_{a}^{b} [f(x)]^2 dx\)

Where \(A\), the area of the region, is \(\int_{0}^{\pi} \[x \sin x\] dx\). Calculating these integrals gives the centroid, which identifies the balancing point on both the x and y axes.

In cases where integrating by hand is troublesome, numerical techniques are often employed to tackle these tasks. This process allows for precise location of the centroid without extensive algebraic manipulation.

Numerical Integration

When a problem involves complex integrals that are difficult to solve analytically, numerical integration becomes a valuable tool. It involves approximating the value of an integral using computational methods instead of symbolic manipulation.

Numerical integration can take several forms:

• Trapezoidal Rule: Uses trapezoids to approximate the area under a curve. It's straightforward and a good starting point for numerical integration.
• Simpson's Rule: Fits parabolic arcs instead of straight lines to better account for curvature in functions.
• Monte Carlo Integration: Utilizes randomness and statistics; useful for high-dimensional integrations.

In the context of this exercise, when dealing with integrals such as \(\pi \int_{0}^{\pi} [x \sin x]^2 dx\) or \(2\pi \int_{0}^{\pi} [x \cdot x \sin x] dx\), numerical methods provide an approximate solution where algebraic integration is not feasible.

Choosing the appropriate method depends on the specific problem conditions, and computational resources available, which can lead to accurate estimations needed for engineering, physics, and other applied sciences.