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Magazine Chapter 7: Problem 62
Find the volume obtained by rotating the region bounded by the curves about the given axis. $$ y=\sin ^{2} x, y=0,0 \leqslant x \leqslant \pi ; \text { about the } x \text { -axis } $$
Short Answer
Expert verified
The volume is \( \frac{3}{8}\pi^2 \).
Step by step solution
01
Understanding the Problem
We need to find the volume generated when the area under the curve \( y = \sin^2(x) \) from \( x = 0 \) to \( x = \pi \) is rotated around the \( x \)-axis. Given \( y = 0 \) is the lower bound, the curve is symmetric around the axis and generates a solid of revolution.
02
Formula for Volume of Revolution
To find the volume of a solid of revolution, we use the disk method. The formula is \[ V = \int_{a}^{b} \pi [f(x)]^2 \, dx \] for rotating around the \( x \)-axis. Here, \( f(x) = \sin^2(x) \), \( a = 0 \), and \( b = \pi \).
03
Set Up the Integral
Substitute \( f(x) \) with \( \sin^2(x) \): \[ V = \int_{0}^{\pi} \pi (\sin^2(x))^2 \, dx \]. Simplifying gives \[ V = \pi \int_{0}^{\pi} \sin^4(x) \, dx \].
04
Simplify the Integral
We will simplify the integral \( \int \sin^4(x) \, dx \) using the power-reduction formula: \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \). Thus \( \sin^4(x) = (\sin^2(x))^2 = \left(\frac{1 - \cos(2x)}{2}\right)^2 = \frac{1}{4}(1 - 2\cos(2x) + \cos^2(2x)) \).
05
Use Another Trigonometric Identity
Further simplify \( \cos^2(2x) \) using the identity: \( \cos^2(2x) = \frac{1 + \cos(4x)}{2} \). Substitute back to get \( \sin^4(x) = \frac{1}{4} (1 - 2\cos(2x) + \frac{1}{2} + \frac{1}{2}\cos(4x)) \) which simplifies to \( \frac{3}{8} - \frac{1}{2}\cos(2x) + \frac{1}{8}\cos(4x) \).
06
Integrate Term-by-Term
Integrate \( \pi \int (\frac{3}{8} - \frac{1}{2}\cos(2x) + \frac{1}{8}\cos(4x)) \, dx \): \- \( \int \frac{3}{8} \, dx = \frac{3}{8}x \) \- \( \int -\frac{1}{2}\cos(2x) \, dx = -\frac{1}{4}\sin(2x) \) \- \( \int \frac{1}{8}\cos(4x) \, dx = \frac{1}{32}\sin(4x) \).
07
Evaluate the Definite Integral
Substitute the bounds 0 to \( \pi \) into \( \frac{3}{8}x - \frac{1}{4}\sin(2x) + \frac{1}{32}\sin(4x) \):- At \( x = \pi \): \( \frac{3}{8}\pi - \frac{1}{4}\sin(2\pi) + \frac{1}{32}\sin(4\pi) = \frac{3}{8}\pi \) since \( \sin(2\pi) = 0 \) and \( \sin(4\pi) = 0 \).- At \( x = 0 \): \( 0 - 0 + 0 = 0 \).Subtract these values to find the integral result \( \frac{3}{8}\pi \).
08
Find the Final Volume
Multiply the result by \( \pi \) to get the volume: \[ V = \pi \times \frac{3}{8}\pi = \frac{3}{8}\pi^2 \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Disk Method
The Disk Method is a technique used in calculus to find the volume of a solid of revolution. This solid is formed by rotating a region around an axis. In our exercise, we are rotating the region under the curve of \( y = \sin^2(x) \) from \( x = 0 \) to \( x = \pi \) around the \( x \)-axis. The method involves slicing the solid perpendicular to the axis of rotation, resulting in a series of disk-shaped slices.
Each disk has a small thickness \( dx \) and a radius equal to the function value \( f(x) \). The volume of a single disk is then given by \( \pi [f(x)]^2 \, dx \).
Each disk has a small thickness \( dx \) and a radius equal to the function value \( f(x) \). The volume of a single disk is then given by \( \pi [f(x)]^2 \, dx \).
- The overall volume is calculated by integrating this expression from \( x = a \) to \( x = b \), where \( a \) and \( b \) are the bounds of the region.
- In our case, \( f(x) = \sin^2(x) \), and the bounds are \( 0 \) to \( \pi \).
- So, the formula for volume becomes \( V = \int_{0}^{\pi} \pi \sin^4(x) \, dx \).
Trigonometric Identities
Trigonometric identities are essential tools when dealing with integrals involving trigonometric functions. For the integral of \( \sin^4(x) \), utilizing these identities is crucial to simplify the expression for easier computation.
When \( \sin^2(x) \) appears, the power-reduction identity \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \) is helpful. By expressing \( \sin^4(x) = (\sin^2(x))^2 \), we get:
When \( \sin^2(x) \) appears, the power-reduction identity \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \) is helpful. By expressing \( \sin^4(x) = (\sin^2(x))^2 \), we get:
- \( \sin^4(x) = \left(\frac{1 - \cos(2x)}{2}\right)^2 \)
- This simplifies to \( \frac{1}{4}(1 - 2\cos(2x) + \cos^2(2x)) \)
- Further simplification using \( \cos^2(2x) = \frac{1 + \cos(4x)}{2} \) leads to a more integrable form.
Definite Integral
The definite integral allows us to calculate the accumulated quantity, such as area or volume, between a specific set of bounds. In the context of finding the volume of a revolution, the definite integral sums up the volumes of all the infinitesimally thin disks from \( x = a \) to \( x = b \).
Once the integral is set up as \( V = \pi \int_{0}^{\pi} \sin^4(x) \, dx \), we need to evaluate it using the bounds \( 0 \) and \( \pi \).
Once the integral is set up as \( V = \pi \int_{0}^{\pi} \sin^4(x) \, dx \), we need to evaluate it using the bounds \( 0 \) and \( \pi \).
- First, we simplify the integrand using trigonometric identities to get an expression that can be integrated piece by piece.
- We perform the integration term by term and evaluate it at the bounds to find the difference.
- This process ensures that we obtain an exact volume for the solid, considering all contributions within the interval of interest.
Power-Reduction Formula
The power-reduction formula is a handy trigonometric identity used to simplify integrals that involve higher powers of sine or cosine. In the problem, \( \sin^4(x) \) initially presents itself as a challenging term to integrate.
Using the power-reduction formula, \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \), we can express \( \sin^4(x) \) in this way:
Using the power-reduction formula, \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \), we can express \( \sin^4(x) \) in this way:
- First, break down \( \sin^4(x) \) to \( \left(\sin^2(x)\right)^2 \).
- Apply the formula: \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \), resulting in \( \sin^4(x) = \left(\frac{1 - \cos(2x)}{2}\right)^2 \).
- This results in a simplified form that is \( \frac{1}{4}(1 - 2\cos(2x) + \cos^2(2x)) \).
