Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the​ x-axis. Recall that cosine squared x equals one half left parenthesis 1 plus cosine 2 x right parenthesis
.
yequalscosine 17 x​, yequals​0, xequals0 
Question content area bottom
Part 1
Set up the integral that gives the volume of the solid. 

To find the volume of the solid generated when the region \( R \) is revolved about the \( x \)-axis, we will use the disk method. The volume of a solid of revolution generated by rotating a function \( y = f(x) \) about the \( x \)-axis from \( x = a \) to \( x = b \) is given by:

\[
V = \pi \int_{a}^{b} [f(x)]^2 \, dx
\]

Given the function \( y = \cos(17x) \), \( y = 0 \), and the bounds \( x = 0 \), the region \( R \) is defined between these curves. We need to determine the bounds for \( x \), which in this case start from \( x = 0 \) and proceed to the point where \( \cos(17x) = 0 \).

### Step-by-Step Setup

1. **Determine the Bounds of Integration:**

   Since \( y = \cos(17x) \) and we are given \( x = 0 \), we look for the first point where \( \cos(17x) = 0 \).

   \[
   \cos(17x) = 0 \implies 17x = \frac{\pi}{2} \implies x = \frac{\pi}{34}
   \]

   So, the region \( R \) is bounded between \( x = 0 \) and \( x = \frac{\pi}{34} \).

2. **Set up the Volume Integral:**

   The volume \( V \) of the solid generated when the region \( R \) is revolved about the \( x \)-axis is:

\[
V = \pi \int_{0}^{\frac{\pi}{34}} [\cos(17x)]^2 \, dx
\]

3. **Simplify the Integrand Using Trigonometric Identity:**

   We are given the identity:

\[
\cos^2(x) = \frac{1}{2} (1 + \cos(2x))
\]

   Substituting \( x = 17x \) in the identity:

\[
\cos^2(17x) = \frac{1}{2} \left(1 + \cos(34x)\right)
\]

   Therefore, the integral becomes:

\[
V = \pi \int_{0}^{\frac{\pi}{34}} \frac{1}{2} \left(1 + \cos(34x)\right) \, dx
\]

4. **Integrate to Find the Volume:**

\[
V = \frac{\pi}{2} \int_{0}^{\frac{\pi}{34}} \left(1 + \cos(34x)\right) \, dx
\]

Now, let's compute this integral.The volume of the solid generated when the region \( R \) is revolved about the \( x \)-axis is:

\[
V = \frac{\pi^2}{68}
\]

Would you like more details on any specific part, or do you have any other questions?

### Related Questions:
1. How would the volume change if the region were revolved around the y-axis instead?
2. Can you compute the volume using the shell method instead of the disk method?
3. What is the volume if the function \( y = \cos(17x) \) is replaced with \( y = \sin(17x) \)?
4. How does the integral change if we revolve the region about a line other than the \( x \)-axis?
5. What are the effects of changing the bounds of integration on the volume?

**Tip:** For trigonometric integrals, using identities can simplify the calculation significantly.

Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the​ x-axis. Recall that cosine squared x equals one half left parenthesis 1 plus cosine 2 x right parenthesis. y = cos(17x), y = 0, x = 0. Set up the integral that gives the volume of the solid.

Explore how to calculate the volume of a solid generated by revolving the region bounded by y = cos(17x), y = 0, and x = 0 around the x-axis using the disk method. This solution involves applying trigonometric identities and integration techniques. Suitable for college-level calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation