The Surface Area of the Solid

Oct 13, 2025 by sorusgentrification.com 61 views

Introduction

In this article, we will explore the concept of a solid formed by the intersection of a parabola and a series of circular disks. The parabola is defined by the equation $x^2 = 4y$ , and the circular disks are perpendicular to the $y$ -axis, with their diameters running from the $y$ -axis to the parabola. We will examine the properties of this solid, including its volume and surface area.

The Parabola

The parabola is a quadratic curve that opens upwards, with its vertex at the origin $(0,0)$ . The equation of the parabola is given by $x^2 = 4y$ , which can be rewritten as $y = \frac{x^2}{4}$ . This equation represents a parabola that opens upwards, with its focus at the point $(0,1)$ .

The Circular Disks

The circular disks are perpendicular to the $y$ -axis, with their diameters running from the $y$ -axis to the parabola. The radius of each disk is equal to the distance from the $y$ -axis to the point of intersection with the parabola. This distance can be calculated using the equation of the parabola, which is $x^2 = 4y$ . By substituting $y = 0$ into this equation, we get $x^2 = 0$ , which implies that $x = 0$ . Therefore, the radius of each disk is equal to the distance from the $y$ -axis to the point $(0,0)$ , which is $0$ .

The Volume of the Solid

The volume of the solid can be calculated by integrating the area of each circular disk with respect to $y$ . The area of each disk is equal to $\pi r^2$ , where $r$ is the radius of the disk. Since the radius of each disk is equal to the distance from the $y$ -axis to the point of intersection with the parabola, we can substitute $r = \sqrt{4y}$ into the equation for the area of each disk. This gives us an area of $\pi (\sqrt{4y})^2 = 4\pi y$ .

To calculate the volume of the solid, we need to integrate the area of each disk with respect to $y$ . This can be done using the following integral:

V = \int_{0}^{2} 4\pi y \, dy

Evaluating this integral, we get:

V = 4\pi \left[\frac{y^2}{2}\right]_{0}^{2} = 4\pi \left(\frac{2^2}{2} - \frac{0^2}{2}\right) = 8\pi

Therefore, the volume of the solid is $8\pi$ cubic units.

The Surface Area of the Solid

The surface area of the solid can be calculated by integrating the area of each circular disk with respect to $y$ . However, we also need to consider the area of the parabolic surface that is not covered by the disks. This area can be calculated using the following integral:

A = \int_{0}^{2} 2\pi x \, dy

Substituting $x = \sqrt{4y}$ into this equation, we get:

A = \int_{0}^{2} 2\pi \sqrt{4y} \, dy

Evaluating this integral, we get:

A = 2\pi \left[\frac{2\sqrt{4y}}{3}\right]_{0}^{2} = 2\pi \left(\frac{2\sqrt{4(2)}}{3} - \frac{2\sqrt{4(0)}}{3}\right) = \frac{16\pi}{3}

Therefore, the surface area of the solid is $\frac{16\pi}{3}$ square units.

Conclusion

In this article, we have explored the concept of a solid formed by the intersection of a parabola and a series of circular disks. We have calculated the volume and surface area of this solid, and have found that the volume is $8\pi$ cubic units and the surface area is $\frac{16\pi}{3}$ square units. This solid is a classic example of a parametric surface, and its properties can be used to model a wide range of real-world phenomena.

References

[1] "Calculus" by Michael Spivak
[2] "Differential Geometry" by John M. Lee
[3] "Parametric Surfaces" by David A. Brannan

Q: What is the equation of the parabola that forms the solid?

A: The equation of the parabola is given by $x^2 = 4y$ , which can be rewritten as $y = \frac{x^2}{4}$ .

Q: What is the radius of each circular disk that forms the solid?

A: The radius of each disk is equal to the distance from the $y$ -axis to the point of intersection with the parabola. This distance can be calculated using the equation of the parabola, which is $x^2 = 4y$ . By substituting $y = 0$ into this equation, we get $x^2 = 0$ , which implies that $x = 0$ . Therefore, the radius of each disk is equal to the distance from the $y$ -axis to the point $(0,0)$ , which is $0$ .

Q: How is the volume of the solid calculated?

A: The volume of the solid can be calculated by integrating the area of each circular disk with respect to $y$ . The area of each disk is equal to $\pi r^2$ , where $r$ is the radius of the disk. Since the radius of each disk is equal to the distance from the $y$ -axis to the point of intersection with the parabola, we can substitute $r = \sqrt{4y}$ into the equation for the area of each disk. This gives us an area of $\pi (\sqrt{4y})^2 = 4\pi y$ .

To calculate the volume of the solid, we need to integrate the area of each disk with respect to $y$ . This can be done using the following integral:

V = \int_{0}^{2} 4\pi y \, dy

Evaluating this integral, we get:

V = 4\pi \left[\frac{y^2}{2}\right]_{0}^{2} = 4\pi \left(\frac{2^2}{2} - \frac{0^2}{2}\right) = 8\pi

Therefore, the volume of the solid is $8\pi$ cubic units.

Q: How is the surface area of the solid calculated?

A: The surface area of the solid can be calculated by integrating the area of each circular disk with respect to $y$ . However, we also need to consider the area of the parabolic surface that is not covered by the disks. This area can be calculated using the following integral:

A = \int_{0}^{2} 2\pi x \, dy

Substituting $x = \sqrt{4y}$ into this equation, we get:

A = \int_{0}^{2} 2\pi \sqrt{4y} \, dy

Evaluating this integral, we get:

A = 2\pi \left[\frac{2\sqrt{4y}}{3}\right]_{0}^{2} = 2\pi \left(\frac{2\sqrt{4(2)}}{3} - \frac{2\sqrt{4(0)}}{3}\right) = \frac{16\pi}{3}

Therefore, the surface area of the solid is $\frac{16\pi}{3}$ square units.

Q: What are some real-world applications of the solid formed by the intersection of a parabola and a series of circular disks?

A: The solid formed by the intersection of a parabola and a series of circular disks has many real-world applications. For example, it can be used to model the shape of a satellite dish or a parabolic mirror. It can also be used to calculate the volume and surface area of a solid object with a parabolic shape.

Q: How can the properties of the solid be used to model real-world phenomena?

A: The properties of the solid can be used to model a wide range of real-world phenomena, including the shape of a satellite dish or a parabolic mirror. It can also be used to calculate the volume and surface area of a solid object with a parabolic shape.

Q: What are some common mistakes to avoid when working with the solid formed by the intersection of a parabola and a series of circular disks?

A: Some common mistakes to avoid when working with the solid formed by the intersection of a parabola and a series of circular disks include:

Failing to consider the radius of each disk when calculating the volume and surface area of the solid.
Failing to integrate the area of each disk with respect to $y$ when calculating the volume and surface area of the solid.
Failing to consider the area of the parabolic surface that is not covered by the disks when calculating the surface area of the solid.

Q: What are some resources for further learning about the solid formed by the intersection of a parabola and a series of circular disks?

A: Some resources for further learning about the solid formed by the intersection of a parabola and a series of circular disks include:

"Calculus" by Michael Spivak
"Differential Geometry" by John M. Lee
"Parametric Surfaces" by David A. Brannan

Q: What are some common applications of the solid formed by the intersection of a parabola and a series of circular disks in engineering and physics?

A: Some common applications of the solid formed by the intersection of a parabola and a series of circular disks in engineering and physics include:

Modeling the shape of a satellite dish or a parabolic mirror.
Calculating the volume and surface area of a solid object with a parabolic shape.
Modeling the behavior of light or sound waves in a parabolic shape.

Q: What are some common challenges when working with the solid formed by the intersection of a parabola and a series of circular disks?

A: Some common challenges when working with the solid formed by the intersection of a parabola and a series of circular disks include:

Calculating the volume and surface area of the solid.
Modeling the behavior of light or sound waves in a parabolic shape.
Ensuring that the solid is properly defined and bounded.