Question

Find the volume generated by revolving the curve \(y=\cos (3 x)\) about the \(x\) -axis, \(0 \leq x \leq \frac{\pi}{36}\).

Short answer

Volume is \(\frac{\pi (\pi + 3)}{72}\) cubic units.

Step-by-step solution

Understand the problem

We need to find the volume of a solid of revolution. The curve given is \(y = \cos(3x)\), and we are rotating it around the \(x\)-axis from \(x = 0\) to \(x = \frac{\pi}{36}\). This can be done using the disk method, which involves integrating the area of circular disks along the \(x\)-axis.

Set up the volume integral using the disk method

The formula for volume using the disk method is \( V = \pi \int_{a}^{b} [f(x)]^2 \, dx \). Here, \(f(x) = \cos(3x)\), \(a = 0\), and \(b = \frac{\pi}{36}\). Therefore, the integral becomes: \[ V = \pi \int_{0}^{\frac{\pi}{36}} [\cos(3x)]^2 \, dx \].

Simplify the integral

The integrand \([\cos(3x)]^2\) can be simplified using the trigonometric identity: \(\cos^2(u) = \frac{1 + \cos(2u)}{2}\). So, the expression becomes:\[ V = \pi \int_{0}^{\frac{\pi}{36}} \frac{1 + \cos(6x)}{2} \, dx \].

Integrate the simplified function

Separate the integral and integrate term-by-term:\[ V = \pi \left( \int_{0}^{\frac{\pi}{36}} \frac{1}{2} \, dx + \int_{0}^{\frac{\pi}{36}} \frac{\cos(6x)}{2} \, dx \right) \]. The first integral is straightforward: \(\int_{0}^{\frac{\pi}{36}} \frac{1}{2} \, dx = \frac{x}{2} \Bigg|_{0}^{\frac{\pi}{36}} = \frac{\pi}{72}\).The second integral requires a substitution: let \(u = 6x\), \(du = 6\,dx\), or \(dx = \frac{du}{6}\). The integral becomes: \(\int \frac{\cos(6x)}{2} \, dx = \frac{1}{12} \sin(6x) \Bigg|_{0}^{\frac{\pi}{36}} = \frac{1}{12} (\sin(\pi/6) - \sin(0)) = \frac{1}{12} \left(\frac{1}{2}\right) = \frac{1}{24}\).

Compute the final volume expression

Now, combine both parts of the integral:\[ V = \pi \left( \frac{\pi}{72} + \frac{1}{24} \right) = \pi \left( \frac{\pi + 3}{72} \right) = \frac{\pi^2 + 3\pi}{72} \].

Simplify the solution

The volume of the solid of revolution is \(\frac{\pi (\pi + 3)}{72}\) cubic units. This is the result of integrating and simplifying the volume expression developed through previous steps.

Key concepts

Disk Method

The disk method is a powerful technique for finding the volume of a solid of revolution. It involves rotating a given curve around an axis and determining the volume by integrating circular disks formed perpendicular to the axis. In this exercise, the curve \( y = \cos(3x) \) is revolved about the \( x \)-axis within the limits \( 0 \leq x \leq \frac{\pi}{36} \). To set up the disk method, visualize slicing the solid into thin, flat disks like a stack of coins. Each disk has a radius equal to the value of the function \( y \) at each \( x \) and a thickness \( dx \). Consequently, the volume of each thin disk is \( \pi [f(x)]^2 dx \), where \( f(x) = \cos(3x) \). Integrating these volumes over the interval gives the entire solid's volume: \[V = \pi \int_{0}^{\frac{\pi}{36}} [\cos(3x)]^2 \, dx\]This method highlights the importance of the function's square since the radius of each disk is directly proportional to the height given by \( f(x) \).

Trigonometric Identities

Trigonometric identities are essential for simplifying expressions involving trigonometric functions. In this problem, specifically, we use the identity for \( \cos^2(u) \): \[\cos^2(u) = \frac{1 + \cos(2u)}{2}\]This identity helps transform the integrand \([\cos(3x)]^2\), making the integral more manageable. Without this simplification, integrating \( [\cos(3x)]^2 \) directly would be more complex. Replacing \( 3x \) with \( u \) in the identity, we get: \[[\cos(3x)]^2 = \frac{1 + \cos(6x)}{2}\]By applying this trigonometric identity, we can express the original square in terms of a sum, separating it into simpler parts that are much easier to integrate.

Volume Integral

A volume integral is a key component in calculating the volume via the disk method. Once the integrand is suitably simplified using trigonometric identities, we set up and evaluate the integral to find the volume of the solid.The integral of the function expressed in terms of \([\cos(3x)]^2\) becomes:\[V = \pi \int_{0}^{\frac{\pi}{36}} \frac{1 + \cos(6x)}{2} \, dx\]This integral breaks down into two simpler parts: - \( \int_{0}^{\frac{\pi}{36}} \frac{1}{2} \, dx \)- \( \int_{0}^{\frac{\pi}{36}} \frac{\cos(6x)}{2} \, dx \)Each part is integrated separately. The first part straightforwardly evaluates the constant, while the second part requires a substitution involving \( u = 6x \) to proceed. These steps ensure accurate evaluation of the volume enclosed by revolution.

Cosine Function

The cosine function \( y = \cos(3x) \) describes the initial curve in this exercise. It is periodic, fluctuating between -1 and 1, and is compressed horizontally by a factor of 3 in comparison to \( y = \cos(x) \). In our problem, it's crucial to understand how this function behaves over the specified interval \( 0 \leq x \leq \frac{\pi}{36} \). This interval covers a very small portion of a complete cosine wave, which is typically \( 2\pi \) long.The behavior of this cosine function directly affects the shape and consequently the volume of the solid of revolution. Each point on the curve contributes to the radius of the disks used in the disk method, making the understanding of cosine behavior pivotal to solving the exercise. Additionally, recognizing the function's symmetry can sometimes simplify calculations involving the integral and revolution operations.