### Problem DescriptionSet up but do not evaluate an integral that represents the volume of the solid of revolution obtained by rotating the region bounded by the curves \( y = 2x + 1 \) and \( y = (x - 1)^2 \) about the y-axis.### ExplanationThis problem involves finding the volume of a solid of revolution. Specifically, the region defined by the given curves is rotated around the y-axis to form a three-dimensional shape. The integral needs to be set up to represent the volume of the resulting solid but does not need to be solved.#### Steps to Set Up the Integral:1. **Identify the curves:** - \( y = 2x + 1 \) - \( y = (x - 1)^2 \)2. **Determine the points of intersection:** Set the equations \( y = 2x + 1 \) and \( y = (x - 1)^2 \) equal to each other to find the points where the curves intersect. \[ 2x + 1 = (x - 1)^2 \] Simplify and solve the quadratic equation. \[ 2x + 1 = x^2 - 2x + 1 \] \[ x^2 - 4x = 0 \] \[ x(x - 4) = 0 \] Therefore, \( x = 0 \) and \( x = 4 \).3. **Set up the integral using the disk method:** Use the disk method to find the volume of the solid of revolution. However, since we are rotating about the y-axis, we need to express \( x \) as a function of \( y \). For \( y = 2x + 1 \): \[ x = \frac{y - 1}{2} \] For \( y = (x - 1)^2 \): \[ x = 1 \pm \sqrt{y} \] Given the range of \( y \) values from the intersection points: When \( x = 0 \), \[ y = 2(0) + 1 = 1 \] When \( x = 4 \), \

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Answered: Set up but do not evaluate an integral…

Set up but do not evaluate an integral that represents the volume of the solid of revolution 2x + 1 and y = (x − 1)² about the obtained by rotating the region bounded by the curves y y-axis.

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)

6th Edition

ISBN:9781337111348

Author:Bruce Crauder, Benny Evans, Alan Noell

Publisher:Bruce Crauder, Benny Evans, Alan Noell

ChapterA: Appendix

SectionA.2: Geometric Constructions

Problem 10P: A soda can has a volume of 25 cubic inches. Let x denote its radius and h its height, both in...

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Question

### Problem Description

Set up but do not evaluate an integral that represents the volume of the solid of revolution obtained by rotating the region bounded by the curves \( y = 2x + 1 \) and \( y = (x - 1)^2 \) about the y-axis.

### Explanation

This problem involves finding the volume of a solid of revolution. Specifically, the region defined by the given curves is rotated around the y-axis to form a three-dimensional shape. The integral needs to be set up to represent the volume of the resulting solid but does not need to be solved.

#### Steps to Set Up the Integral:

1. **Identify the curves:**
- \( y = 2x + 1 \)
- \( y = (x - 1)^2 \)

2. **Determine the points of intersection:**
Set the equations \( y = 2x + 1 \) and \( y = (x - 1)^2 \) equal to each other to find the points where the curves intersect.

\[
2x + 1 = (x - 1)^2
\]

Simplify and solve the quadratic equation.

\[
2x + 1 = x^2 - 2x + 1
\]
\[
x^2 - 4x = 0
\]
\[
x(x - 4) = 0
\]

Therefore, \( x = 0 \) and \( x = 4 \).

3. **Set up the integral using the disk method:**
Use the disk method to find the volume of the solid of revolution. However, since we are rotating about the y-axis, we need to express \( x \) as a function of \( y \).

For \( y = 2x + 1 \):
\[
x = \frac{y - 1}{2}
\]

For \( y = (x - 1)^2 \):
\[
x = 1 \pm \sqrt{y}
\]

Given the range of \( y \) values from the intersection points:
When \( x = 0 \),
\[
y = 2(0) + 1 = 1
\]

When \( x = 4 \),
\

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