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Chapter 6: Problem 25

Write an integral for the area of the surface generated by revolving the curve \(y=\cos x,-\pi / 2 \leq x \leq \pi / 2,\) about the \(x\) -axis. In Section 8.5 we will see how to evaluate such integrals.

Short Answer

Expert verified
The integral for the surface area is \(\int_{-\pi/2}^{\pi/2} 2\pi \cos x \, dx\).

Step by step solution

01

Understand the Problem

The problem is asking us to write an integral expression for the area of a surface of revolution. Specifically, it involves revolving the function \(y = \cos x\) around the \(x\)-axis over the interval \(-\pi/2 \leq x \leq \pi/2\).
02

Formula for Surface Area of Revolution

The formula to calculate the surface area \(A\) of a curve \(y = f(x)\) revolved around the \(x\)-axis from \(x = a\) to \(x = b\) is: \[ A = \int_{a}^{b} 2\pi y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]
03

Differentiate the Function

Calculate \(\frac{dy}{dx}\) for \(y = \cos x\). The derivative of \(\cos x\) is \(-\sin x\), so: \[ \frac{dy}{dx} = -\sin x \]
04

Substitute into the Surface Area Formula

Substitute \(y = \cos x\) and \(\frac{dy}{dx} = -\sin x\) into the surface area formula:\[ A = \int_{-\pi/2}^{\pi/2} 2\pi \cos x \sqrt{1 + (-\sin x)^2} \, dx \]
05

Simplify the Expression Inside the Integral

Simplify the expression under the square root:\[ 1 + (-\sin x)^2 = 1 + \sin^2 x = \cos^2 x + \sin^2 x \] By the Pythagorean identity, this equals 1.So, the integral becomes:\[ A = \int_{-\pi/2}^{\pi/2} 2\pi \cos x \cdot 1 \, dx = \int_{-\pi/2}^{\pi/2} 2\pi \cos x \, dx \]

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Key Concepts

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

Integral Calculus
Integral calculus is an essential branch of mathematics. It helps us calculate areas, volumes, and other values that arise from integrating functions. When considering surfaces of revolution, integral calculus is a handy tool.
The area of a surface generated by revolving a curve around an axis can often be determined using integrals. This involves integrating over the interval of the curve.
In the context of our problem, the surface is created by revolving the function \(y = \cos x\) around the \(x\)-axis. We use the bounds from \(-\pi/2\) to \(\pi/2\) since these define the portion of the curve we are interested in.
  • The basic idea is to "add up" infinitely small rings or disks that make up the surface.
  • You use the formula \( A = \int_{a}^{b} 2\pi y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \) and customize it for your function and interval.
Integral calculus thus offers a structured method to "add up" these tiny rings and derive the total surface area.
Differentiation
Differentiation is a vital process in calculus that involves finding the derivative of a function. The derivative represents the rate of change of the function with respect to a variable.
In the given exercise, to use the surface area formula, we must calculate the derivative of the function \( y = \cos x \). Differentiation rules tell us that the derivative of \( \cos x \) is \( -\sin x \).
  • Not only does this derivative tell us about the slope of the tangent to the curve at any point, it is crucial for determining the arc length aspect in the surface area formula.
  • This differentiation step transforms the surface area formula into a workable integral.
By calculating \( \frac{dy}{dx} \), we enrich the surface area equation, allowing us to move toward integrating it to find the total area.
Pythagorean Identity
The Pythagorean identity is a fundamental trigonometric principle, representing the key relationship between sine and cosine.
The identity is expressed as \( \sin^2 x + \cos^2 x = 1 \). This is particularly useful in simplifying expressions in integration and differentiation processes.
In the surface area of revolution problem, this identity comes into play when simplifying the expression under the square root in the integral:
  • The expression \( 1 + (-\sin x)^2 \) simplifies to \( 1 + \sin^2 x \).
  • Using the Pythagorean identity, \( 1 + \sin^2 x \) converts to \( \cos^2 x + \sin^2 x \), simplifying to just \( 1 \).
This simplification is crucial as it reduces the complexity of the integral, letting us effectively handle and compute the surface area. Employing the Pythagorean identity streamlines solving the problem, making the integral more manageable.

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Most popular questions from this chapter

The region in the first quadrant that is bounded above by the curve \(y=1 / \sqrt{x},\) on the left by the line \(x=1 / 4,\) and below by the line \(y=1\) is revolved about the \(y\) -axis to generate a solid. Find the volume of the solid by a. the washer method. b. the shell method.

Find the center of mass of a thin plate covering the region bounded below by the parabola \(y=x^{2}\) and above by the line \(y=x\) if the plate's density at the point \((x, y)\) is \(\delta(x)=12 x .\)

Lifting an elevator cable An electric elevator with a motor at the top has a multistrand cable weighing 4.5 \(\mathrm{lb} / \mathrm{ft}\) . When the car is at the first floor, 180 \(\mathrm{ft}\) of cable are paid out, and effectively 0 \(\mathrm{ft}\) are out when the car is at the top floor. How much work does the motor do just lifting the cable when it takes the car from the first floor to the top?

Leaky sandbag A bag of sand originally weighing 144 lb was lifted at a constant rate. As it rose, sand also leaked out at a constant rate. The sand was half gone by the time the bag had been lifted to 18 \(\mathrm{ft}\) . How much work was done lifting the sand this far? (Neglect the weight of the bag and lifting equipment.)

Compute the volume of the solid generated by revolving the triangular region bounded by the lines \(2 y=x+4, y=x,\) and \(x=0\) about a. the \(x\) -axis using the washer method. b. the \(y\) -axis using the shell method. c. the line \(x=4\) using the shell method. d. the line \(y=8\) using the washer method.
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