Question

To find: The Volume of the solid which is obtained on rotating the region bounded by the given curves about the specified line.

Short answer

The volume of the solid is \(\frac{5}{{14}}\pi \).

Step-by-step solution

Given information

The function is \(y = {x^3},y = 0,x = 1\).

The concept of volume by cylindrical shell method

Definition of Volume by cylindrical shell method:

Consider that the solid\(S\)is obtained by rotation of the region under the curve\(y = f(x)\)about the\(y\)-axis from\(a\)to\(b\)is\(V = \int_a^b 2 \pi xf(x)dx\).

Calculate the volume

The bounded curves of a region are assumed to be \(y = {x^3},y = 0,x = 1\).

The specific line around which the region is rotated is\(y = 1\)which is parallel to\(x\)-axis.

Obtain the value of\(x\)as follows:

\(\begin{aligned}{}{x^3} = 0\\x = 0\end{aligned}\)

Here, the shell radius is\(1 - y\)and the shell height is\(1 - {y^{\frac{1}{3}}}\).

The region lies between\(y = 0\), and\(y = 1\), so by definition of cylindrical shell, the volume becomes

\(\begin{aligned}{}V = \int_0^1 2 \pi (1 - y)\left( {1 - {y^{\frac{1}{3}}}} \right)dy\\ = 2\pi \int_0^1 {\left( {1 - {y^{\frac{1}{3}}} - y + {y^{\frac{4}{3}}}} \right)} dy\\ = 2\pi \left( {y - \frac{{3{y^{\frac{4}{3}}}}}{4} - \frac{{{y^2}}}{2} + \frac{{3{y^{\frac{7}{3}}}}}{7}} \right)_0^1\\ = 2\pi \left( {1 - \frac{3}{4} - \frac{1}{2} + \frac{3}{7}} \right)\end{aligned}\)

On further simplifications,

\(\begin{aligned}{}V = 2\pi \left( {\frac{{28 - 21 - 14 + 12}}{{28}}} \right)\\ = 2\pi \left( {\frac{5}{{28}}} \right)\\ = \frac{{10}}{{28}}\pi \\ = \frac{5}{{14}}\pi \end{aligned}\)

Therefore, the volume of the solid is \(\frac{5}{{14}}\pi \).