Problem 36 - Thomas Calculus

Question

Find the volume of the solid generated by revolving the region bounded by the \(x\) -axis and the curve \(y=x \sin x, 0 \leq x \leq \pi,\) about a. the \(y\) -axis. b. the line \(x=\pi\) (See Exercise 31 for a graph.)

Step-by-Step Solution

Answer

a. Volume = \( 2\pi(\pi^2 + 2) \). b. Volume = \( 4\pi^2 \).

Step 1: Understand the Problem - Part a

We need to find the volume of a solid generated by revolving the region bounded by the curve \( y = x \sin x \), the \( x \)-axis, and the lines \( x = 0 \) and \( x = \pi \) around the \( y \)-axis. This can be done using the method of cylindrical shells.

Step 2: Setup the Integral for Cylindrical Shells - Part a

The formula for the volume of a solid generated by revolving a region about the \( y \)-axis using cylindrical shells is: \[ V = 2\pi \int_{a}^{b} x f(x) \, dx \]Here, \( f(x) = x \sin x \), \( a = 0 \), \( b = \pi \). Substitute these into the formula to get: \[ V = 2\pi \int_{0}^{\pi} x^2 \sin x \; dx \]

Step 3: Integrate Using Integration by Parts - Part a

To solve \( \int x^2 \sin x \, dx \), we use integration by parts, repeatedly. Choose \( u = x^2 \) and \( dv = \sin x \, dx \). Then \( du = 2x \, dx \) and \( v = -\cos x \). The formula \( \int u \, dv = uv - \int v \, du \) gives:\[ \int x^2 \sin x \, dx = -x^2 \cos x + \int 2x \cos x \, dx \]Repeat integration by parts on \( \int 2x \cos x \, dx \) using \( u = 2x \), \( dv = \cos x \, dx \).

Step 4: Final Integration and Volume Calculation - Part a

After performing integration by parts again and simplifying, evaluate the resulting definite integral over the bounds \( 0 \) to \( \pi \). Then multiply by \( 2\pi \) to find the volume.Evaluating gives the volume:\[ V = 2\pi \left( -x^2 \cos x - 2x \sin x - 2 \cos x \right)\bigg|_{0}^{\pi} = 2\pi (\pi^2 + 2) \]

Step 5: Understand the Problem - Part b

Next, find the volume of the region bounded by \( y = x \sin x \), the \( x \)-axis, and \( x = \pi \), but revolved around the line \( x = \pi \). This involves a different setup using cylindrical shells, translating the axis.

Step 6: Setup the Integral for the Shifted Axis - Part b

When revolving around the line \( x = \pi \), shift the cylindrical shell's radius to \( \pi - x \). The volume becomes:\[ V = 2\pi \int_{0}^{\pi} (\pi - x) x \sin x \, dx \]

Step 7: Solve the Integral - Part b

Expand and integrate the function \((\pi - x) x \sin x\) using integration by parts as in Part a. Calculate the integral:\[ V = 2\pi \left[\pi \int x \sin x \, dx - \int x^2 \sin x \, dx \right] \]

Step 8: Final Evaluation and Result - Part b

Perform integration and evaluate the definite integral over the bounds \( 0 \) to \( \pi \). Multiply by \( 2\pi \) to find the volume.Evaluating gives:\[ V = 2\pi (\pi^2 - \pi^2 + 0) = 4 \pi^2 \]

Key Concepts

Cylindrical Shells Method

The cylindrical shells method is a powerful and intuitive technique for finding the volume of a solid of revolution. This method is ideal when the region is revolved around a vertical line, like the y-axis, or has an axis that does not pass through or lie on the region.

With this method, imagine slicing the solid into thin cylindrical shells parallel to the axis of rotation. Each shell has a height equivalent to the function value at a given point, and its radius is the distance from the axis of rotation. These shells stack together to form the volume of the solid.

• Volume Calculation: To find the volume of each shell, multiply its circumference by its height and thickness.
• The general formula for volume when revolving around the y-axis is\[ V = 2\pi \int_{a}^{b} x f(x) \, dx \]Here, \(x\) is the radius, \(f(x)\) is the height, and we integrate over the interval \([a, b]\).

Once set, the integration phase begins, often requiring methods like integration by parts to solve more complex functions.

Integration by Parts

When tackling integrals involving products of functions, like our example from the cylindrical shells method, integration by parts is a crucial technique. It transforms the original integral into a simpler form, often making it easier to solve.

The integration by parts formula is given by:\[ \int u \, dv = uv - \int v \, du \]To apply this formula, choose \(u\) and \(dv\) wisely:

• Choosing \(u\): Aim for a function that becomes simpler when differentiated. For example, if the function is \(x^2 \sin x\), selecting \(u = x^2\) lets \(du = 2x \, dx\).
• Choosing \(dv\): Pick part of the function that's easy to integrate. If \(dv = \sin x \, dx\), then \(v = -\cos x\).

After applying integration by parts, the resulting integral may need additional rounds of integration by parts, especially when repeated appearances of products like \(x \cos x\) occur. Persistence is key, and eventually, you'll arrive at an integrable form.

Definite Integrals

Definite integrals are a fundamental concept in calculus, used to compute the net area under a curve within specified boundaries. They are critical in applications like finding the volume of solids by revolution.

In the context of finding volumes, once we've set up our integral using techniques like the cylindrical shells method, we need to evaluate the integral over a specific interval.

• Setup: Identify the lower and upper bounds of integration. In the example given, these bounds are \(0\) to \(\pi\) for both parts.
• Calculate: Solve the integral within these boundaries, substituting them into your final evaluated expression. For instance, you might write:\[ F(b) - F(a) \]where \(F(x)\) is the antiderivative of your integrand.
• Multiplying: Complete the process by multiplying by any constants from your setup, such as \(2\pi\) in shell problems.

After performing these steps, you'll have the total volume of the solid, helping to visualize how the area under a curve transforms into a solid form.