Find The Volume Of The Solid Whose Base Is Bounded By Y = 1 − X 2 Y = 1 - \frac{x}{2} Y = 1 − 2 X ​ , Y = X 2 − 1 Y = \frac{x}{2} - 1 Y = 2 X ​ − 1 , And The Y-axis. The Cross Sections Of The Solid, Taken Perpendicular To The X-axis, Are Equilateral Triangles. Round Your Answer To

by ADMIN 282 views

Introduction

In this article, we will explore the problem of finding the volume of a solid whose base is bounded by the lines y=1−x2y = 1 - \frac{x}{2}, y=x2−1y = \frac{x}{2} - 1, and the y-axis. The cross sections of the solid, taken perpendicular to the x-axis, are equilateral triangles. This problem requires us to use the method of disks and the properties of equilateral triangles to find the volume of the solid.

Understanding the Problem

To begin, let's visualize the problem. We have a solid whose base is bounded by three lines: y=1−x2y = 1 - \frac{x}{2}, y=x2−1y = \frac{x}{2} - 1, and the y-axis. The y-axis is the line x=0x = 0. The cross sections of the solid, taken perpendicular to the x-axis, are equilateral triangles. This means that if we were to slice the solid at any point, the resulting cross section would be an equilateral triangle.

Finding the Base of the Solid

The base of the solid is bounded by the lines y=1−x2y = 1 - \frac{x}{2}, y=x2−1y = \frac{x}{2} - 1, and the y-axis. To find the base of the solid, we need to find the points where these lines intersect. We can do this by setting the equations equal to each other and solving for x.

Setting y=1−x2y = 1 - \frac{x}{2} and y=x2−1y = \frac{x}{2} - 1 equal to each other, we get:

1−x2=x2−11 - \frac{x}{2} = \frac{x}{2} - 1

Solving for x, we get:

x=0x = 0

This means that the lines intersect at the point (0, 0). Therefore, the base of the solid is bounded by the lines y=1−x2y = 1 - \frac{x}{2}, y=x2−1y = \frac{x}{2} - 1, and the y-axis.

Finding the Height of the Solid

The height of the solid is the distance between the two lines y=1−x2y = 1 - \frac{x}{2} and y=x2−1y = \frac{x}{2} - 1. To find the height, we need to find the difference between the two lines.

The difference between the two lines is:

(1−x2)−(x2−1)\left(1 - \frac{x}{2}\right) - \left(\frac{x}{2} - 1\right)

Simplifying, we get:

2−x2 - x

This is the height of the solid.

Finding the Volume of the Solid

To find the volume of the solid, we need to use the method of disks. The method of disks states that the volume of a solid is equal to the integral of the area of the cross sections with respect to x.

In this case, the area of the cross sections is the area of an equilateral triangle. The area of an equilateral triangle is given by:

A=34s2A = \frac{\sqrt{3}}{4} s^2

where s is the side length of the triangle.

The side length of the triangle is equal to the height of the solid, which is 2−x2 - x. Therefore, the area of the cross sections is:

A=34(2−x)2A = \frac{\sqrt{3}}{4} (2 - x)^2

The volume of the solid is equal to the integral of the area of the cross sections with respect to x. Therefore, we can write:

V=∫0234(2−x)2dxV = \int_{0}^{2} \frac{\sqrt{3}}{4} (2 - x)^2 dx

Evaluating the integral, we get:

V=34∫02(4−4x+x2)dxV = \frac{\sqrt{3}}{4} \int_{0}^{2} (4 - 4x + x^2) dx

V=34[4x−2x2+x33]02V = \frac{\sqrt{3}}{4} \left[4x - 2x^2 + \frac{x^3}{3}\right]_0^2

V=34[8−8+83]V = \frac{\sqrt{3}}{4} \left[8 - 8 + \frac{8}{3}\right]

V=34[83]V = \frac{\sqrt{3}}{4} \left[\frac{8}{3}\right]

V=233V = \frac{2\sqrt{3}}{3}

Therefore, the volume of the solid is 233\frac{2\sqrt{3}}{3} cubic units.

Conclusion

In this article, we found the volume of a solid whose base is bounded by the lines y=1−x2y = 1 - \frac{x}{2}, y=x2−1y = \frac{x}{2} - 1, and the y-axis. The cross sections of the solid, taken perpendicular to the x-axis, are equilateral triangles. We used the method of disks and the properties of equilateral triangles to find the volume of the solid. The volume of the solid is 233\frac{2\sqrt{3}}{3} cubic units.

References

  • [1] "Calculus: Early Transcendentals" by James Stewart
  • [2] "Calculus" by Michael Spivak
  • [3] "Mathematics for the Nonmathematician" by Morris Kline

Introduction

In our previous article, we explored the problem of finding the volume of a solid whose base is bounded by the lines y=1−x2y = 1 - \frac{x}{2}, y=x2−1y = \frac{x}{2} - 1, and the y-axis. The cross sections of the solid, taken perpendicular to the x-axis, are equilateral triangles. In this article, we will answer some of the most frequently asked questions about this problem.

Q: What is the method of disks?

A: The method of disks is a technique used to find the volume of a solid. It states that the volume of a solid is equal to the integral of the area of the cross sections with respect to x.

Q: What is the area of an equilateral triangle?

A: The area of an equilateral triangle is given by the formula:

A=34s2A = \frac{\sqrt{3}}{4} s^2

where s is the side length of the triangle.

Q: How do we find the side length of the equilateral triangle?

A: The side length of the equilateral triangle is equal to the height of the solid, which is 2−x2 - x.

Q: How do we find the volume of the solid?

A: To find the volume of the solid, we need to use the method of disks. We integrate the area of the cross sections with respect to x, and the result is the volume of the solid.

Q: What is the volume of the solid?

A: The volume of the solid is 233\frac{2\sqrt{3}}{3} cubic units.

Q: Why do we need to use the method of disks?

A: We need to use the method of disks because the solid is not a simple shape, and we cannot find its volume using other methods. The method of disks allows us to break down the solid into smaller parts and find its volume by integrating the area of the cross sections.

Q: Can we use other methods to find the volume of the solid?

A: Yes, we can use other methods to find the volume of the solid, such as the method of washers or the method of cylindrical shells. However, the method of disks is the most straightforward and easiest to use in this case.

Q: What are the limitations of the method of disks?

A: The method of disks has some limitations. It is only applicable to solids whose cross sections are circular or can be approximated as circular. It is also not applicable to solids with irregular shapes or complex boundaries.

Q: Can we use the method of disks to find the volume of other solids?

A: Yes, we can use the method of disks to find the volume of other solids, as long as the cross sections are circular or can be approximated as circular.

Conclusion

In this article, we answered some of the most frequently asked questions about finding the volume of a solid with equilateral triangle cross sections. We discussed the method of disks, the area of an equilateral triangle, and how to find the volume of the solid. We also discussed the limitations of the method of disks and how it can be used to find the volume of other solids.

References

  • [1] "Calculus: Early Transcendentals" by James Stewart
  • [2] "Calculus" by Michael Spivak
  • [3] "Mathematics for the Nonmathematician" by Morris Kline