Find the volume of the solid formed by revolving the region bounded by the graphs of y = x³, x = 2, and y = 1 about the y-axis.

Solution:

Given, y = x³

x = 2 and y = 1 about the y-axis.

We have to find the volume of the solid formed by revolving the region bounded by the given graphs.

Using the shell method,

The height of the shell is determined by the vertical distance between the curve y = x³ and the line y = 1.

The radius of each shell is determined by the value of x, which can only range from 1 to 2.

The height is determined exactly by x³- 1 at this x.

Multiply by 2π to get the surface area of this thin cylindrical shell.

Next, integrate along the interval [1, 2] to get the volume.

We have,

Volume = \(2\pi \int_{1}^{2}x(x^{3}-1)\, dx\\=2\pi \int_{1}^{2}(x^{4}-x)\, dx\\=2\pi\left [ \frac{x^{5}}{5}-\frac{x^{2}}{2} \right ]_{1}^{2}\\=2\pi [(\frac{(2)^{5}}{5}-\frac{(2)^{2}}{2})-(\frac{(1)^{5}}{5}-\frac{(1)^{2}}{2})]\)

\(\\=2\pi[(\frac{32}{5}-\frac{4}{2})-(\frac{1}{5}-\frac{1}{2})]\\=2\pi[\frac{32}{5}-2-\frac{1}{5}+\frac{1}{2}]\\=2\pi [\frac{32}{5}-\frac{1}{5}+\frac{1}{2}-2]\\=2\pi [\frac{31}{5}-\frac{3}{2}]\\=2\pi [\frac{(62-15)}{10}]\\=2\pi [\frac{47}{10}]\)

\(=\pi [\frac{47}{5}]\)

Therefore, the volume of the solid is \(=\pi [\frac{47}{5}]\)cubic units.