Volume of Revolution by Disk Method: Calculating the Solid Generated by the Curve y x^3

The volume of a solid of revolution generated by rotating a region around an axis is a classic application in calculus. In this article, we explore how to find the volume of the solid formed by revolving the region bounded by the curve y x3, the x-axis, and the lines x 1 and x 2 about the x-axis using the disk method. This technique is particularly useful and is widely accepted in various fields, including engineering and physics.

Understanding the Disk Method

The disk method involves cutting the solid into infinitesimally thin cylindrical disks. Each disk has a small thickness (dx) and a radius given by the function being revolved, y x3. The volume of each disk is given by (dV pi y^2 , dx), and integrating this over the entire region gives the total volume of the solid.

Step-by-Step Calculation

Given the function (y x^3) and the bounds (x 1) and (x 2), the volume (V) can be expressed as:

[V pi int_{1}^{2} (x^3)^2 , dx]

Simplifying this, we get:

[V pi int_{1}^{2} x^6 , dx]

Next, we integrate:

[int x^6 , dx frac{x^7}{7}]

Now we evaluate this from (1) to (2):

[V pi left[ frac{x^7}{7} right]_{1}^{2}]

Substituting the values, we obtain:

[V pi left( frac{2^7}{7} - frac{1^7}{7} right)]

[V pi left( frac{128}{7} - frac{1}{7} right)]

[V pi left( frac{127}{7} right)]

The final volume is:

[V frac{127pi}{7}]

Alternative Methods

There are several other methods to calculate the volume of the solid of revolution. Here, we discuss three different approaches to solve the same problem.

shell Method

A shell approach involves considering a thin shell created by revolving an infinitesimal element about the x-axis. For (y x^3), the shell has a thickness (dy), a length (L 2 - y^{frac{1}{3}}), and a circumference (C 2pi y). The volume of each shell is:

[dV 2pi y cdot L , dy]

Simplifying this integral:

[V 2pi int_{0}^{8} y left(2 - y^{frac{1}{3}}right) , dy]

[V 2pi int_{0}^{8} left(2y - y^{frac{4}{3}}right) , dy]

Evaluating this integral:

[V 2pi left[ y^2 - frac{3}{7} y^{frac{7}{3}} right]_{0}^{8}]

[V 2pi left( 64 - frac{3}{7} cdot 2^7 right)]

[V 2pi left( 64 - frac{384}{7} right)]

[V frac{128pi}{7}]

Disc Method

The disc method involves summing the volumes of infinitesimal discs of thickness (dx), each with an area (A pi y^2). The volume is given by:

[V pi int_{0}^{2} x^6 , dx]

This integral evaluates to:

[V pi left[ frac{1}{7} x^7 right]_{0}^{2}]

[V pi left( frac{1}{7} cdot 2^7 right)]

[V frac{128pi}{7}]

Second Theorem of Pappus

The second theorem of Pappus states that the volume is given by:

[V 2pi bar{y} A]

First, calculate the area (A):

[A int_{0}^{2} x^3 , dx]

[A frac{1}{4} x^4 Big|_{0}^{2}]

[A frac{1}{4} cdot 2^4]

[A 4]

Then, calculate (bar{y}):

[bar{y} frac{1}{A} int_{0}^{2} frac{x^3}{2} x^3 , dx]

[bar{y} frac{1}{2A} int_{0}^{2} x^6 , dx]

[bar{y} frac{1}{8} cdot frac{x^7}{7} Big|_{0}^{2}]

[bar{y} frac{16}{7}]

Finally, the volume using Pappus's theorem is:

[V 2pi bar{y} A]

[V 2pi cdot frac{16}{7} cdot 4]

[V frac{128pi}{7}]

Conclusion

This article demonstrates that although there are multiple methods to calculate the volume of a solid of revolution, the disk method offers a straightforward and clear solution. The second theorem of Pappus provides an alternative route when the centroid's location is known. Each method has its advantages and can be used depending on the problem's complexity and the available information.