The base of a three-dimensional figure is bound by the line x = -2y - 2 on the interval [-4, -1]. Vertical cross sections that are perpendicular to the y-axis are squares. Find the volume of the figure. 2 1 X -21 1 2 3 4 5 6 7 8 ŠŅ♡+44p ņ op -2+ -3 t
The base of a three-dimensional figure is bound by the line x = -2y - 2 on the interval [-4, -1]. Vertical cross sections that are perpendicular to the y-axis are squares. Find the volume of the figure. 2 1 X -21 1 2 3 4 5 6 7 8 ŠŅ♡+44p ņ op -2+ -3 t
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Volume Calculation for a Three-Dimensional Figure**
The base of a three-dimensional figure is defined by the line \( x = -2y - 2 \) over the interval \([-4, -1]\). The vertical cross-sections perpendicular to the y-axis are square shapes. We need to determine the volume of this figure.
**Step-by-Step Solution:**
1. **Interpret the boundary line equation:**
The given line \( x = -2y - 2 \) represents a linear boundary of the base in the \(xy\)-plane.
2. **Determine the side length of the squares:**
Since the vertical cross-sections are squares and given the equation \( x = -2y - 2 \), the side length \( s \) of each square at any y-coordinate over the range \([-4, -1]\) can be derived from the equation of the line.
3. **Volume integration setup:**
To find the volume, we integrate the area of the squares along the interval of \( y \) from \(-4\) to \(-1\). The area of a square is \( s^2 \).
First, express \( s \) in terms of \( y \):
\[
s = |-2y - 2|
\]
Since \( y \) is in the interval \([-4, -1\]):
\[
s = -2y - 2
\]
4. **Calculate the volume using integration:**
We need to integrate the area of the squares:
\[
V = \int_{-4}^{-1} s^2 \, dy
\]
Substitute \( s = -2y - 2 \):
\[
V = \int_{-4}^{-1} (-2y - 2)^2 \, dy
\]
Simplify the expression inside the integral:
\[
(-2y - 2)^2 = 4y^2 + 8y + 4
\]
Thus, the integral becomes:
\[
V = \int_{-4}^{-1} (4y^2 + 8y + 4) \, dy
\]
Compute the integral:
\[
V = \left[ \frac{4y^3}{3} +](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6bc5b610-4c42-40dc-b81a-6ff9e1fb2857%2F9c5748de-f88c-4de4-bf36-0a81d756eedd%2Fm44wqij_processed.png&w=3840&q=75)
Transcribed Image Text:**Volume Calculation for a Three-Dimensional Figure**
The base of a three-dimensional figure is defined by the line \( x = -2y - 2 \) over the interval \([-4, -1]\). The vertical cross-sections perpendicular to the y-axis are square shapes. We need to determine the volume of this figure.
**Step-by-Step Solution:**
1. **Interpret the boundary line equation:**
The given line \( x = -2y - 2 \) represents a linear boundary of the base in the \(xy\)-plane.
2. **Determine the side length of the squares:**
Since the vertical cross-sections are squares and given the equation \( x = -2y - 2 \), the side length \( s \) of each square at any y-coordinate over the range \([-4, -1]\) can be derived from the equation of the line.
3. **Volume integration setup:**
To find the volume, we integrate the area of the squares along the interval of \( y \) from \(-4\) to \(-1\). The area of a square is \( s^2 \).
First, express \( s \) in terms of \( y \):
\[
s = |-2y - 2|
\]
Since \( y \) is in the interval \([-4, -1\]):
\[
s = -2y - 2
\]
4. **Calculate the volume using integration:**
We need to integrate the area of the squares:
\[
V = \int_{-4}^{-1} s^2 \, dy
\]
Substitute \( s = -2y - 2 \):
\[
V = \int_{-4}^{-1} (-2y - 2)^2 \, dy
\]
Simplify the expression inside the integral:
\[
(-2y - 2)^2 = 4y^2 + 8y + 4
\]
Thus, the integral becomes:
\[
V = \int_{-4}^{-1} (4y^2 + 8y + 4) \, dy
\]
Compute the integral:
\[
V = \left[ \frac{4y^3}{3} +
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