# Find the volume of the frustrum of the right circular cone generated by rotating the line y = x +...

## Question:

Find the volume of the frustrum of the right circular cone generated by rotating the line {eq}y = x + 1 {/eq} about the x-axis between {eq}x = 0 {/eq} and {eq}x = 3 {/eq}.

## Volume of the Frustrum:

To find the volume of the frustrum, use the formula of the frustrum.

First, find the area of the cross-section, then find the volume.

The formula of the volume of the frustrum is

{eq}V = \int_{a}^{b} A\left ( x \right ) dx {/eq}.

Here, A\left ( x \right ) is area of the cross-section.

The formula is {eq}A\left ( x \right ) = \pi y^{2 {/eq}.

Given: {eq}y=x+1 {/eq}

The x-axis between x=0 and x = 4.

Use disk method about the x-axis.

Area of cross section is {eq}A\left ( x \right ) = \pi \left ( x+1 \right )^{2} {/eq}

The volume of the solid {eq}V = \int_{a}^{b} A\left ( x \right ) dx {/eq}.

{eq}= \pi \int_{0}^{3} \left ( x^2 + 2x + 1 \right ) dx {/eq}

{eq}= \pi \left(\int _0^3 x^2 dx + \int _0^3 2x dx + \int _0^3 1 dx \right) {/eq}

{eq}= \pi \left [ \left[\frac{x^{2+1}}{2+1}\right]_{0}^{3} + 2\left[\frac{x^{1+1}}{1+1}\right]_{0}^{3} + \left[x\right]_{0}^{3} \right ] {/eq}

{eq}= \pi \left [ 9 + 9 + 3 \right ] {/eq}

{eq}= 21\pi {/eq}

Hence, the volume of the frustum is {eq}21\pi {/eq}. 