To solve the integral for the volume, let's continue from where we left off:

The integral we need to solve is:
\[
V = \int_{0}^{\frac{\pi}{6}} \left( 14 \sin^2 3\theta + \frac{8 \sin^3 3\theta}{3} \right) \, d\theta
\]

Let's break it into two parts:
\[
V = \int_{0}^{\frac{\pi}{6}} 14 \sin^2 3\theta \, d\theta + \int_{0}^{\frac{\pi}{6}} \frac{8 \sin^3 3\theta}{3} \, d\theta
\]

1. **First Integral**:
\[
\int_{0}^{\frac{\pi}{6}} 14 \sin^2 3\theta \, d\theta
\]
Using the identity \(\sin^2 x = \frac{1 - \cos 2x}{2}\), we get:
\[
\sin^2 3\theta = \frac{1 - \cos 6\theta}{2}
\]
So,
\[
14 \sin^2 3\theta = 14 \cdot \frac{1 - \cos 6\theta}{2} = 7 (1 - \cos 6\theta)
\]
Thus,
\[
\int_{0}^{\frac{\pi}{6}} 14 \sin^2 3\theta \, d\theta = \int_{0}^{\frac{\pi}{6}} 7 (1 - \cos 6\theta) \, d\theta
\]
\[
= 7 \left[ \theta - \frac{\sin 6\theta}{6} \right]_{0}^{\frac{\pi}{6}}
\]
Evaluating the integral:
\[
= 7 \left( \frac{\pi}{6} - \frac{\sin (\pi)}{6} - (0 - 0) \right)
\]
\[
= 7 \left( \frac{\pi}{6} \right)
\]
\[
= \frac{7\pi}{6}
\]

2. **Second Integral**:
\[
\int_{0}^{\frac{\pi}{6}} \frac{8 \sin^3 3\theta}{3} \, d\theta
\]
Using the identity \(\sin^3 x = \frac{3 \sin x - \sin 3x}{4}\), we get:
\[
\sin^3 3\theta = \frac{3 \sin 3\theta - \sin 9\theta}{4}
\]
So,
\[
\frac{8 \sin^3 3\theta}{3} = \frac{8}{3} \cdot \frac{3 \sin 3\theta - \sin 9\theta}{4} = 2 \left( \sin 3\theta - \frac{\sin 9\theta}{3} \right)
\]
Thus,
\[
\int_{0}^{\frac{\pi}{6}} \frac{8 \sin^3 3\theta}{3} \, d\theta = \int_{0}^{\frac{\pi}{6}} 2 \left( \sin 3\theta - \frac{\sin 9\theta}{3} \right) \, d\theta
\]
\[
= 2 \left[ -\frac{\cos 3\theta}{3} + \frac{\cos 9\theta}{27} \right]_{0}^{\frac{\pi}{6}}
\]
Evaluating the integral:
\[
= 2 \left( -\frac{\cos \frac{\pi}{2}}{3} + \frac{\cos \frac{3\pi}{2}}{27} - \left( -\frac{\cos 0}{3} + \frac{\cos 0}{27} \right) \right)
\]
\[
= 2 \left( 0 + 0 - \left( -\frac{1}{3} + \frac{1}{27} \right) \right)
\]
\[
= 2 \left( \frac{1}{3} - \frac{1}{27} \right)
\]
\[
= 2 \left( \frac{9 - 1}{27} \right)
\]
\[
= 2 \left( \frac{8}{27} \right)
\]
\[
= \frac{16}{27}
\]

Combining the results:
\[
V = \frac{7\pi}{6} + \frac{16}{27}
\]

To combine these fractions:
\[
\frac{7\pi}{6} + \frac{16}{27} = \frac{7\pi \cdot 27 + 16 \cdot 6}{6 \cdot 27}
\]
\[
= \frac{189\pi + 96}{162}
\]

Therefore, the volume \(V\) is:
\[
V = \frac{189\pi + 96}{162}
\]

Would you like further details or have any questions? Here are some related questions you might find useful:

1. How do you verify the bounds of integration in cylindrical coordinates?
2. What are the steps to convert a volume integral into cylindrical coordinates?
3. What trigonometric identities are most useful for integrating products of trigonometric functions?
4. How can numerical integration techniques be applied to complex integrals?
5. What is the geometric interpretation of the solid bounded by these surfaces?
6. How do cylindrical and spherical coordinates compare for volume integration problems?
7. What role do symmetry and bounds play in simplifying volume integrals?
8. How can software tools assist in solving complex integrals?

**Tip:** Always look for symmetries and patterns in the integrand to simplify the computation.

Learn how to solve the integral for volume calculation using trigonometric identities. This problem involves integrating sine squared and sine cubed functions over a specified interval. Suitable for college-level students and enthusiasts of calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation