### Setting Up the Integral for Volume of a Solid of RevolutionTo find the volume of the solid generated by revolving the region bounded by $ y = x^2 - 3 $ and $ y = 2x $ about the line $ y = -4 $:Below are several integral expressions; choose the correct one based on the given problem:1. \[ \pi \int_{-1}^{3} \left((x^2 - 3)^2 - (2x)^2 \right) \, dx \]2. \[ \pi \int_{-1}^{3} \left( -x^2 + 1 \right)^2 - \left(2x - 4 \right)^2 \, dx \]3. \[ \pi \int_{-1}^{3} \left(2x + 4 \right)^2 - \left( x^2 + 1 \right)^2 \, dx \]4. \[ \pi \int_{-1}^{3} \left( x^2 + 1 \right)^2 - \left( 2x + 4 \right)^2 \, dx \]The correct integral expression is highlighted:\[\pi \int_{-1}^{3} \left((x^2 - 3)^2 - (2x)^2 \right) \, dx\]This setup involves using the washer method for finding the volume of a solid of revolution. The radius to the outer function $ y = x^2 - 3 $ and the radius to the inner function $ y = 2x $ come into play when revolving around the line $ y = -4 $.**Explanation:**- The integral is set up from the intersection points of the two functions, observed within the limits $ x = -1 $ and $ x = 3 $.- Each function is transformed to align with the horizontal axis shifted so that the revolution is about $ y = -4 $.- The volume calculation uses the difference of squares of the radii, reinforced by the given choices.This process emphasizes understanding integrals in the context of solid of revolution volumes, highlighting critical steps such as transformation, limits, and application of the washer method.

Answered: Image /qna-images/answer/6f22c078-410d-4cb1-bc37-1be56afe0783.jpg

Set up but do not evaluate the integral used to find the volume of the solid generated by revolving the region bounded by y = x² – 3 and y = 2x about the liney = -4. °z(* - 3)° - 2) d.x :서(-2 + 1).-(2x-4)? dx L:( 2x + 4 )* – (x² + 1)² dx - :(x* + 1)° – (2x + 4ỷ° dz 4)² dx -

Set up but do not evaluate the integral used to find the volume of the solid generated by revolving the region bounded by y = x² – 3 and y = 2x about the liney = -4. °z(* - 3)° - 2) d.x :서(-2 + 1).-(2x-4)? dx L:( 2x + 4 )* – (x² + 1)² dx - :(x* + 1)° – (2x + 4ỷ° dz 4)² dx -

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...

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Question

### Setting Up the Integral for Volume of a Solid of Revolution

To find the volume of the solid generated by revolving the region bounded by $ y = x^2 - 3 $ and $ y = 2x $ about the line $ y = -4 $:

Below are several integral expressions; choose the correct one based on the given problem:

1. \[
\pi \int_{-1}^{3} \left((x^2 - 3)^2 - (2x)^2 \right) \, dx
\]

2. \[
\pi \int_{-1}^{3} \left( -x^2 + 1 \right)^2 - \left(2x - 4 \right)^2 \, dx
\]

3. \[
\pi \int_{-1}^{3} \left(2x + 4 \right)^2 - \left( x^2 + 1 \right)^2 \, dx
\]

4. \[
\pi \int_{-1}^{3} \left( x^2 + 1 \right)^2 - \left( 2x + 4 \right)^2 \, dx
\]

The correct integral expression is highlighted:

\[
\pi \int_{-1}^{3} \left((x^2 - 3)^2 - (2x)^2 \right) \, dx
\]

This setup involves using the washer method for finding the volume of a solid of revolution. The radius to the outer function $ y = x^2 - 3 $ and the radius to the inner function $ y = 2x $ come into play when revolving around the line $ y = -4 $.

**Explanation:**
- The integral is set up from the intersection points of the two functions, observed within the limits $ x = -1 $ and $ x = 3 $.
- Each function is transformed to align with the horizontal axis shifted so that the revolution is about $ y = -4 $.
- The volume calculation uses the difference of squares of the radii, reinforced by the given choices.

This process emphasizes understanding integrals in the context of solid of revolution volumes, highlighting critical steps such as transformation, limits, and application of the washer method.

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