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Chapter 6: Problem 14

Find the volume of the solid that results when the region enclosed by the given curves is revolved about the \(x\) -axis. $$y=\sin x, y=\cos x, x=0, x=\pi / 4$$ [Hint: Use the identity \(\left.\cos 2 x=\cos ^{2} x-\sin ^{2} x .\right]\)

Short Answer

Expert verified
The volume is \(\frac{\pi}{2}\).

Step by step solution

01

Understanding the Region

First, identify the region enclosed by the given curves. We have the functions \(y = \sin x\) and \(y = \cos x\) and the vertical lines \(x = 0\) and \(x = \pi / 4\). This region can be described as the area between these two curves from \(x = 0\) to \(x = \pi / 4\).
02

Setting Up the Volume Integral

We will use the washer method to find the volume. When revolved around the \(x\)-axis, the volume \(V\) of the solid is given by the integral of the difference of squares of the outer radius and inner radius: \[ V = \pi \int_0^{\pi/4} [(\cos x)^2 - (\sin x)^2] \, dx \] Here, \(\cos x\) is the outer radius and \(\sin x\) is the inner radius.
03

Simplify the Integrand Using Trigonometric Identity

Apply the given identity \(\cos 2x = \cos^2 x - \sin^2 x\) to simplify the integrand:\[ V = \pi \int_0^{\pi/4} \cos 2x \, dx \]
04

Integrate with Respect to x

Calculate the integral:First find the antiderivative of \(\cos 2x\):\[ \int \cos 2x \, dx = \frac{1}{2}\sin 2x + C \]Now evaluate the definite integral:\[ V = \pi \left[ \frac{1}{2}\sin 2x \right]_0^{\pi/4} \]
05

Evaluate the Definite Integral

Substitute the bounds into the antiderivative:\[ V = \pi \left( \frac{1}{2} \sin \frac{\pi}{2} - \frac{1}{2} \sin 0 \right) \]Since \(\sin \frac{\pi}{2} = 1\) and \(\sin 0 = 0\), we simplify this to:\[ V = \pi \left( \frac{1}{2} \times 1 - 0 \right) = \frac{\pi}{2} \]
06

Conclusion

The volume of the solid that results from revolving the given region around the \(x\)-axis is \(\frac{\pi}{2}\).

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Key Concepts

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

Washer Method
The washer method is a technique used in calculus to find the volume of a solid of revolution. This method is particularly useful when dealing with regions that, when revolved, create solids with hollowed-out sections, similar to a donut or "washer." To apply the washer method, follow these steps:
  • Identify the outer and inner curves that bound the region you are revolving. These are typically two functions, such as in the case of the equation from your exercise: \( y = \cos x \) and \( y = \sin x \).
  • The volume \( V \) of the solid is computed by integrating the difference of the squares of the outer radius and the inner radius over the specified interval. The formula is: \[ V = \pi \int_a^b \left[(\text{outer radius})^2 - (\text{inner radius})^2\right] \, dx \]
  • For your problem, the outer radius is \( \cos x \) and the inner radius is \( \sin x \), creating the integral: \[ V = \pi \int_0^{\pi/4} \left[(\cos x)^2 - (\sin x)^2\right] \, dx \]
This method effectively calculates the volume by considering every little "slice" perpendicular to the axis of rotation as a washer, adding up the volumes of these slices to get the total volume.
Trigonometric Integration
Trigonometric integration involves the integration of trigonometric functions, often using identities to simplify the process. When tasked with integrating a function involving terms like \( \cos^2 x \) and \( \sin^2 x \), trigonometric identities can be quite helpful. In this exercise, the given identity \( \cos 2x = \cos^2 x - \sin^2 x \) simplifies the integrand:
  • Replace \( \cos^2 x - \sin^2 x \) with \( \cos 2x \) in your integral. This simplifies the resulting expression and makes the integration process easier.
  • The integral now becomes: \[ \pi \int_0^{\pi/4} \cos 2x \, dx \]
Using identities not only simplifies the calculations but also reduces the complexity of evaluating definite integrals with trigonometric functions. It is a critical skill in calculus to recognize and apply such identities to solve problems efficiently.
Definite Integrals
A definite integral involves computing the integral of a function over a specific interval, yielding the net area under the curve between two points on the \(x\)-axis. In the exercise above, this concept is employed to find the volume of a solid of revolution:
  • First, find the antiderivative of the integrand, which in this case is \( \cos 2x \). The antiderivative is \( \frac{1}{2} \sin 2x + C \).
  • Next, evaluate this antiderivative from \( x = 0 \) to \( x = \pi/4 \). For definite integrals, the constant \( C \) is not needed. \[ \pi \left[ \frac{1}{2} \sin 2x \right]_0^{\pi/4} \]
  • Substitute the bounds into your antiderivative: \[ \pi \left( \frac{1}{2} \sin \frac{\pi}{2} - \frac{1}{2} \sin 0 \right) \]
Understanding how definite integrals work allows you to compute actual quantities, such as volumes or areas, derived from the accumulation of infinitesimally small values between set limits. In this context, the calculation gives the volume of the solid as \( \frac{\pi}{2} \), representing the entire region when revolved around the \(x\)-axis.

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