Question

How do you find the volume of region bounded by graphs of `y = x^2` and `y = sqrt x` about the x-axis?

Answer 1

`color(blue)(pi/3 "cubic units.")`

Explanation:

From the graph we can see that the volume we seek is between the two functions. In order to find this, we must find the volume of revolution of `f(x)=sqrt(x)` and subtract the volume of revolution of `f(x)=x^2`. This is shown as the shaded area.

First we need to find the upper and lower bounds. We know the lower bound is `0` since `f(x)=sqrt(x)` is undefined for `x<0`. The upper bound is where the functions intersect:

`:.`

`x^2=sqrt(x)`

`x^2/x^(1/2)=1`

`x^(3/2)=1`

Squaring:

`x^3=1`

`x=root(3)(1)=1`

Volume of `bb(f(x)=sqrt(x))`:

`pi int_(0)^(1)(x^(1/2))=pi[2/3x^(3/2)]_(0)^(1)`

`=pi{[2/3x^(3/2)]^(1)-[2/3x^(3/2)]_(0)}`

Plugging in upper and lower bounds:

`=pi{[2/3(1)^(3/2)]^(1)-[2/3(0)^(3/2)]_(0)}=(2pi)/3` cubic units

Volume of `bb(f(x)=x^2)`

`pi int_(0)^(1)(x^2)=pi[1/3x^3]_(0)^(1)`

`=pi{[[1/3x^3]^(1)-[1/3x^3]_(0)}`

Plugging in upper and lower bounds:

`=pi{[[1/3(1)^3]^(1)-[1/3(0)^3]_(0)}=pi/3` cubic units.

Required volume is:

`(2pi)/3-pi/3=` `color(blue)(pi/3 "cubic units.")`

Volume of revolution: