Question

Let R be the region enclosed by the graphs of `y=(64x)^(1/4)` and `y=x`. How do you find the volume of the solid generated when region R is revolved about the x-axis?

Answer 1

`(64pi)/3`

Explanation:

First, plot the two graphs, shading the bounded region between them.


Second, recall the formula for solid of revolution about the `x`-axis

`int_(x_0)^(x_f)pi("outer radius")^2-pi("inner radius")2dx`

The outer radius is the height from the `x`-axis to the outer most graph, or `y=(64x)^(1/4)`

`int_(x_0)^(x_f)pi((64x)^(1/4))^2-pi("inner radius")^2dx`

The inner radius is the height from the `x`-axis to the inner most graph, or `y=x`

`int_(x_0)^(x_f)pi((64x)^(1/4))^2-pi(x)^2dx`

The lower limit of integration is the smallest `x`-value where the two curves meet, or `x_0=0`. The upper limit of integration is the largest `x`-value where the two curves meet, or `x_f=4`

`int_(0)^(4)pi((64x)^(1/4))^2-pi(x)^2dx`

`int_(0)^(4)pi(64x)^(1/2)-pix^2dx`

`int_(0)^(4)pi8x^(1/2)-pix^2dx`

`int_(0)^(4)pi8x^(1/2)dx-int_(0)^(4)pix^2dx`

`8piint_(0)^(4)x^(1/2)dx-piint_(0)^(4)x^2dx`

`16/3pi[x^(3/2)]_(0)^(4)-pi/3[x^3]_(0)^(4)`

`16/3pi(4)^(3/2)-pi/3(4)^3=(64pi)/3`