Set up (but do not evaluate) a definite integral for the volume V of the solid in the figure. The base of the solid is a disk with radius 3. Parallel cross sections perpendicular to the base are squares. :
Set up (but do not evaluate) a definite integral for the volume V of the solid in the figure. The base of the solid is a disk with radius 3. Parallel cross sections perpendicular to the base are squares. :
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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![_Set up (but do not evaluate) a definite integral for the volume \( V \) of the solid in the figure. The base of the solid is a disk with radius 3. Parallel cross sections perpendicular to the base are squares._
**Diagram Explanation:**
The diagram shows a three-dimensional solid with a circular base. The base is a disk with a radius of 3 units. The cross sections, perpendicular to the circular base, are square-shaped. The solid appears to have a smooth, continuous structure, with colors transitioning from blue at the top to orange at the bottom, indicating height or depth.
To find the volume of this solid, you can use a definite integral. The key is to express the area of each square cross section as a function of its position along the radius of the disk. Since each cross section is square, its side length will vary based on its distance from the center of the base.
**Setting Up the Integral:**
1. **Radius of the base (disk):** \( r = 3 \)
2. **Equation of the disk:** \( x^2 + y^2 \leq 9 \)
For a cross section at a distance \( x \) from the center of the disk, the side of the square, \( s \), will be twice the height from the \( x\)-axis, which is \( s = 2\sqrt{9 - x^2} \).
3. **Area of square cross section:** \( A(x) = \left(2\sqrt{9 - x^2}\right)^2 = 4(9 - x^2) \)
4. **Volume integral setup:**
\[
V = \int_{-3}^{3} 4(9 - x^2) \, dx
\]
This integral will produce the volume of the solid by summing up the areas of all the infinitesimally thin square cross sections from \( x = -3 \) to \( x = 3 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F11b54239-8e64-41df-bbac-e376c02f6bfa%2F16154caf-4e28-4d8a-9288-3e8fca9a0ab3%2F739wsz_processed.png&w=3840&q=75)
Transcribed Image Text:_Set up (but do not evaluate) a definite integral for the volume \( V \) of the solid in the figure. The base of the solid is a disk with radius 3. Parallel cross sections perpendicular to the base are squares._
**Diagram Explanation:**
The diagram shows a three-dimensional solid with a circular base. The base is a disk with a radius of 3 units. The cross sections, perpendicular to the circular base, are square-shaped. The solid appears to have a smooth, continuous structure, with colors transitioning from blue at the top to orange at the bottom, indicating height or depth.
To find the volume of this solid, you can use a definite integral. The key is to express the area of each square cross section as a function of its position along the radius of the disk. Since each cross section is square, its side length will vary based on its distance from the center of the base.
**Setting Up the Integral:**
1. **Radius of the base (disk):** \( r = 3 \)
2. **Equation of the disk:** \( x^2 + y^2 \leq 9 \)
For a cross section at a distance \( x \) from the center of the disk, the side of the square, \( s \), will be twice the height from the \( x\)-axis, which is \( s = 2\sqrt{9 - x^2} \).
3. **Area of square cross section:** \( A(x) = \left(2\sqrt{9 - x^2}\right)^2 = 4(9 - x^2) \)
4. **Volume integral setup:**
\[
V = \int_{-3}^{3} 4(9 - x^2) \, dx
\]
This integral will produce the volume of the solid by summing up the areas of all the infinitesimally thin square cross sections from \( x = -3 \) to \( x = 3 \).
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