Use the disk method to find the volume of the solid generated when the region bounded by y = 8 sinx and y = 0, for 0≤x≤, is revolved about the x-axis. (Recall that sin ²x=(1- cos 2x).)

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
icon
Related questions
Question

Help! I know how to set up the integral but I'm having trouble finding f(x)^2

**Volume of Solids Using the Disk Method**

To find the volume of the solid generated when the region bounded by \( y = 8 \sin x \) and \( y = 0 \), for \( 0 \leq x \leq \pi \), is revolved about the x-axis, we will use the disk method. Recall that \(\sin^2 x = \frac{1}{2}(1 - \cos 2x)\).

**Disk Method Formula:**

The volume \(V\) of the solid of revolution formed by revolving the region around the x-axis can be found using the integral:
\[ V = \pi \int_{a}^{b} [f(x)]^2 dx \]

**Step-by-Step Process:**

1. **Identify the functions and limits of integration:**
   - The functions are \( y = 8 \sin x \) and \( y = 0 \).
   - The limits are from \( x = 0 \) to \( x = \pi \).
   
2. **Set up the integral:**
   \[ V = \pi \int_{0}^{\pi} (8 \sin x)^2 dx \]
   
3. **Simplify the integrand:**
   \[ V = \pi \int_{0}^{\pi} 64 \sin^2 x dx \]

4. **Use the identity to rewrite \(\sin^2 x\):**
   \[ \sin^2 x = \frac{1}{2} (1 - \cos 2x) \]
   So,
   \[ 64 \sin^2 x = 64 \left( \frac{1}{2} (1 - \cos 2x) \right) = 32 (1 - \cos 2x) \]

5. **Substitute and integrate:**
   \[ V = \pi \int_{0}^{\pi} 32 (1 - \cos 2x) dx \]
   \[ V = 32 \pi \int_{0}^{\pi} (1 - \cos 2x) dx \]

6. **Evaluate the integral:**
   Break it into two:
   \[ V = 32 \pi \left[ \int_{0}^{\pi} 1 \, dx - \int_{0}^{\pi} \cos 2x
Transcribed Image Text:**Volume of Solids Using the Disk Method** To find the volume of the solid generated when the region bounded by \( y = 8 \sin x \) and \( y = 0 \), for \( 0 \leq x \leq \pi \), is revolved about the x-axis, we will use the disk method. Recall that \(\sin^2 x = \frac{1}{2}(1 - \cos 2x)\). **Disk Method Formula:** The volume \(V\) of the solid of revolution formed by revolving the region around the x-axis can be found using the integral: \[ V = \pi \int_{a}^{b} [f(x)]^2 dx \] **Step-by-Step Process:** 1. **Identify the functions and limits of integration:** - The functions are \( y = 8 \sin x \) and \( y = 0 \). - The limits are from \( x = 0 \) to \( x = \pi \). 2. **Set up the integral:** \[ V = \pi \int_{0}^{\pi} (8 \sin x)^2 dx \] 3. **Simplify the integrand:** \[ V = \pi \int_{0}^{\pi} 64 \sin^2 x dx \] 4. **Use the identity to rewrite \(\sin^2 x\):** \[ \sin^2 x = \frac{1}{2} (1 - \cos 2x) \] So, \[ 64 \sin^2 x = 64 \left( \frac{1}{2} (1 - \cos 2x) \right) = 32 (1 - \cos 2x) \] 5. **Substitute and integrate:** \[ V = \pi \int_{0}^{\pi} 32 (1 - \cos 2x) dx \] \[ V = 32 \pi \int_{0}^{\pi} (1 - \cos 2x) dx \] 6. **Evaluate the integral:** Break it into two: \[ V = 32 \pi \left[ \int_{0}^{\pi} 1 \, dx - \int_{0}^{\pi} \cos 2x
Expert Solution
steps

Step by step

Solved in 2 steps with 2 images

Blurred answer
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Precalculus
Precalculus
Calculus
ISBN:
9780135189405
Author:
Michael Sullivan
Publisher:
PEARSON
Calculus: Early Transcendental Functions
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning