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Chapter 7: Problem 77

Use any method to find the volume of the solid generated when the region enclosed by the curves is revolved about the \(y\) -axis. $$ y=\cos x, y=0, x=0, x=\pi / 2 $$

Short Answer

Expert verified
The volume is \(\pi^2 + 2\pi\).

Step by step solution

01

Understand the Problem

We are given a region bounded by the curve \(y = \cos x\), the line \(y = 0\), and the vertical lines \(x = 0\) and \(x = \frac{\pi}{2}\). This region is revolved around the \(y\)-axis. We need to find the volume of the resulting solid. The method of cylindrical shells is suitable here since we revolve around the \(y\)-axis.
02

Set Up the Integral Using the Shell Method

The formula for the volume using the shell method is \( V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx \). For the given problem, the radius of the shell is \(x\) and the height is \(\cos x\). The limits of integration are from \(x = 0\) to \(x = \frac{\pi}{2}\). The volume integral is therefore \( V = 2\pi \int_{0}^{\pi/2} x \cdot \cos x \, dx \).
03

Solve the Integral

To solve the integral, we perform integration by parts. Let \( u = x \) so \( du = dx \), and let \( dv = \cos x \, dx \) so \( v = \sin x \). The integration by parts formula, \( \int u \, dv = uv - \int v \, du \), gives us: \[V = 2\pi \left[ x \sin x \ Big |_{0}^{\pi/2} - \int_{0}^{\pi/2} \sin x \, dx \right]\]Calculating this yields:\[V = 2\pi \left[ \frac{\pi}{2} \times 1 - \left(-\cos x \right) \Big |_{0}^{\pi/2} \right]\]\[V = 2\pi \left[ \frac{\pi}{2} + ( \cos 0 - \cos \frac{\pi}{2}) \right]\]\[V = 2\pi \left[ \frac{\pi}{2} + (1 - 0) \right] = 2\pi \times \left( \frac{\pi}{2} + 1 \right)\]
04

Simplify the Result

Simplify the expression obtained from the integral:\[V = 2\pi \times \left( \frac{\pi}{2} + 1 \right) = \pi^2 + 2\pi\]
05

Write the Final Answer

The volume of the solid generated when the region is revolved around the \(y\)-axis is \( \pi^2 + 2\pi \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

cylindrical shells method
When dealing with volumes of solids generated by revolving a region around an axis, the cylindrical shells method can be very helpful. This method works particularly well when rotating around the y-axis. Imagine cutting the solid into thin vertical slices. Each slice resembles a hollow cylinder, or shell, which is why it's called the cylindrical shells method.
  • The formula involves integrating the circumference of each shell against its height and thickness: \[ V = 2\pi \int_{{a}}^{{b}} x \cdot f(x) \, dx \]
  • Here, \(x\) represents the radius of the shell (distance from the y-axis), and \(f(x)\) is the height.
This method simplifies many problems because you don't need to solve for \(x\) in terms of \(y\) or manipulate y-values if you're revolving around the y-axis. It's a straightforward approach where each component of your integral has a clear geometric interpretation:
  • \(2\pi x\): The circumference of the shell
  • \(f(x)\): The height of the shell
integration by parts
Integration by parts is a useful technique for solving integrals, especially when the integral is the product of two functions. It's like the product rule for differentiation but in reverse. The formula is: \[ \int u \, dv = uv - \int v \, du \] Let's break it down:
  • Choose \(u\) and \(dv\) such that \(du\) and \(v\) are easier to integrate.
  • Differentiate \(u\) to find \(du\).
  • Integrate \(dv\) to find \(v\).

Next, plug these into the formula to complete the integration. For instance, in our exercised problem, choosing \(u = x\) and \(dv = \cos x \, dx\) allows us to reduce the integral into simpler terms.
Using integration by parts can sometimes be a trial-and-error process, but is invaluable for integrals involving products of algebraic and trigonometric functions.
trigonometric integration
Trigonometric integration involves integrating functions composed of trigonometric identities. These integrals often arise in revolving problems due to the nature of periodic functions. To integrate trigonometric functions, we often employ specific identities or substitutions. Here are some tips and strategies:
  • Familiarize yourself with basic identities such as \[ \sin^2x + \cos^2x = 1 \]. These can help simplify integrals before they are solved.
  • Consider using identities like double angle or product-to-sum when dealing with multiples of trigonometric functions.
  • Substitute \( u = \sin x \) or \( u = \cos x \) if parts of the function match these derivatives.

For instance, transforming \(\int \cos x \, dx\) to its antiderivative, which is \(\sin x + C\), requires knowing the basic relationship between these trigonometric functions.
These strategies are a powerful toolkit in calculus to manage complex integrals arising from geometric or physical situations.

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Most popular questions from this chapter

Engineers want to construct a straight and level road \(600 \mathrm{ft}\) long and \(75 \mathrm{ft}\) wide by making a vertical cut through an intervening hill (see the accompanying figure). Heights of the hill above the centerline of the proposed road, as obtained at various points from a contour map of the region, are shown in the accompanying figure. To estimate the construction costs, the engineers need to know the volume of earth that must be removed. Approximate this volume, rounded to the nearest cubic foot. [Hint: First set up an integral for the cross-sectional area of the cut along the centerline of the road, then assume that the height of the hill does not vary between the centerline and edges of the road.] $$ \begin{array}{cc} \hline \begin{array}{l} \text { HORIZONTAL } \\ \text { DISTANCE } x \text { (ft) } \end{array} & \begin{array}{c} \text { HEIGHT } \\ h(\mathrm{ft}) \end{array} \\ \hline 0 & 0 \\ 100 & 7 \\ 200 & 16 \\ 300 & 24 \\ 400 & 25 \\ 500 & 16 \\ 600 & 0 \end{array} $$

(a) Complete the square, make an appropriate \(u\) substitution, and then use the Endpaper Integral Table to evaluate the integral. (b) If you have a CAS, use it to evaluate the integral (no substitution or square completion), and then confirm that the result is equivalent to that in part (a). $$ \int \frac{x}{x^{2}+6 x+13} d x $$

(a) Make an appropriate \(u\) -substitution of the form \(u=x^{1 / n}\) or \(u=(x+a)^{1 / n}\), and then evaluate the integral. (b) If you have a CAS, use it to evaluate the integral, and then confirm that the result is equivalent to the one that you found in part (a). $$ \int \frac{1+\sqrt{x}}{1-\sqrt{x}} d x $$

Let \(R\) be the region to the right of \(x=1\) that is bounded by the \(x\) -axis and the curve \(y=1 / x .\) When this region is revolved about the \(x\) -axis it generates a solid whose surface is known as Gabriel's Horn (for reasons that should be clear from the accompanying figure). Show that the solid has a finite volume but its surface has an infinite area. [Note: It has been suggested that if one could saturate the interior of the solid with paint and allow it to seep through to the surface, then one could paint an infinite surface with a finite amount of paint! What do you think?]

The region bounded below by the \(x\) -axis and above by the portion of \(y=\sin x\) from \(x=0\) to \(x=\pi\) is revolved about the \(x\) -axis. Find the volume of the resulting solid.
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