Question

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 \(\frac{{32}}{{15}}\pi \).

Step-by-step solution

Given information

The function is \(y = \sqrt x ,y = 0,x = 1\).

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

It is given that the curves that bound a region are \(y = \sqrt x ,y = 0,x = 1\).

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

With\(x = 1\), the value of\(y\)becomes\(y = 1\).

Here, the shell radius is\(1 + x\)and the shell height is\(\sqrt x \).

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

\(\begin{array}{c}V = \int_0^1 2 \pi (1 + x)(\sqrt x )dx\\ = 2\pi \int_0^1 {\left( {{x^{\frac{1}{2}}} + {x^{\frac{3}{2}}}} \right)} dx\\ = 2\pi \left[ {\frac{{2{x^{\frac{3}{2}}}}}{3} + \frac{{3{x^{\frac{5}{2}}}}}{5}} \right]_0^1\end{array}\)

On further simplification,

\(\begin{array}{c}V = 2\pi \left( {\frac{{2(1)\frac{3}{2}}}{3} + \frac{{2{{(1)}^{\frac{5}{2}}}}}{5}} \right)\\ = 2\pi \left( {\frac{{10 + 6}}{{15}}} \right)\\ = \frac{{32\pi }}{{15}}\end{array}\)

Therefore, the volume of the solid is \(\frac{{32}}{{15}}\pi \).