Question

Hi everyone, Use the washer method to find the volume of the solid generated when the region bounded by y equals 3x and y equals x^2 squared is revolved about the​ x-axis. What would be the Volume of the region?

the bounds are b. 2 and a.0

I keep getting 54pi. Am I doing something incorrectly?

Answer 1

`"Volume "= 88/5pi " cubic units"`

Explanation:

If we look at the two functions:

`f(x)=3x` and `f(x)=x^2`

In the interval:

`[0,2]` ie the lower and upper bounds, we notice that:

`f(x)=3x` has greater values than `f(x)=x^2` here. Therefore `3x` is above `x^2`.

In order to find the volume of the region between these two when rotated around the x axis, we need to find the volume of `f(x)=3x` and then subtract the volume of `f(x)=x^2`. This is because our area that we rotate doesn't meet the x axis:

Volume of `f(x)=3x`

Since `3x` is the radius of a cylinder or shell to be rotated and the thickness of each shell is `deltax`, then the volume of each shell will be:

`pir^2*deltax`

The sum of these shells will be found using the integral:

`piint_(0)^(2)(3x)^2dx`

`piint_(0)^(2)(9x^2)dx=pi{[3x^3]^(2)-[3x^3]_(0)}`

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

Plugging in upper and lower bounds:

`V=pi{[3(2)^3]^(2)-[3(0)^3]_(0)}`

`V=pi{24]^(2)-[0]_(0)}=24pi`

We do the same for `f(x)=x^2`

`piint_(0)^(2)(x^2)^2dx=pi{[1/5x^5]^(2)-[1/5x^5]_(0)}`

`V=pi{[1/5x^5]^(2)-[1/5x^5]_(0)}`

Plugging in upper and lower bounds:

`V=pi{[1/5(2)^5]^(2)-[1/5(0)^5]_(0)}`

`V=pi{[32/5]^(2)-[0]_(0)}=pi(32/5)`

Subtracting volume of `f(x)=x^2` from volume of `f(x)=3x`

`24pi-32/5pi=88/5pi`

`"Volume "= 88/5pi " cubic units"`

PLOT: