Question

The volume of the solid with the help of cylindrical shells.

Short answer

The volume of Torus is \(4\int_0^r 2 \pi (R + x)\sqrt {{r^2} - {x^2}} dx\).

Step-by-step solution

Given information

We have given two radius R and r.

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 two radius are \(R\) and \(r\).

Consider the vertical line is at the center of the circle with radius\(r\)and the axis of rotation is at the center of the circle with radius\(R\).

Let the equation of the circle with radius\(r\)be\(y = \sqrt {{r^2} - {x^2}} \).

Take the quarter of the circle with radius\(r\)such the shell radius is\(R + x\)and the shell height is\(r\)

The region of integration is\(x = 0\)to\(x = r\).

The volume of the solid when quarter is rotated about the axis of rotation becomes

\(V = \int_0^r 2 \pi (R + x)\sqrt {{r^2} - {x^2}} dx\)

Multiply this volume by 4, the volume of the Torus is obtained.

Therefore, the volume of Torus is \(4\int_0^r 2 \pi (R + x)\sqrt {{r^2} - {x^2}} dx\).