For the region bounded by `x=0` [vertical line] , `y=0` [horizontal line] and `y =3(2-x)` [slant line], it will have inverted cone when revolved about the y-axis.

Please see the attached image file for the figure.

When bounded plane region revolves about an axis, we can use a rectangular strip. When the rectangular strip is adjacent or perpendicular to the axis of revolution, we may apply DISK method.

This is an integration application for volume with the formula as:

`V = int_a^b A(x) dx` or `V = int_a^b A(y) dy.`

In this problem, we consider `A =pir^2` for a disk area where `r = x= f(y)` and thickness of the strip` = dy` .

For the ` r` , we consider the length of the strip `= x_2-x_1`

For the `x_(2)` , we consider the right bound based on the slant line: `y = 3(2-x)` .

Rearrange this into: `x= 2-y/3` for `x_(2)` .

For the `x_(1)` , we consider the left bound based on boundary line x=0.

Then `r =2-y/3-0`

`r =2-y/3` .

Plug-in `r =2-y/3` on `A =pir^2` , we get:

`A =pi(2-y/3)^2`

`A =pi(2-y/3)*(2-y/3)`

`A =pi(4-4y/3+ y^2/9)`

Based on the bounded region revolve about the y-axis, the thickness of the rectangular strip is `dy` and boundary values: `y=0` to `y =6` .

We can apply: `V = int_a^b A(y)dy` .

`V = int_0^6pi(4-4y/3+ y^2/9)dy.`

Apply basic integration formula: `int c* f dx = c int f dx` where `c` is constant

and `int (u+v+w) dy=int (u) dy+ int (v) dy+int (w) dy` .

`V = pi int_0^6 (4-4y/3+ y^2/9)dy`

`V = pi [int_0^6 4 dy - int_0^6 4y/3dy+ int_0^6 y^2/9dy]`

`V = pi [4int_0^6 dy - 4/3 int_0^6 ydy+ 1/9int_0^6 y^2/9dy]`

Apply power rule for integration: `int y^n dy= y^(n+1)/(n+1)`

`V = pi[4 y - 4/3* y^2/2+ 1/9*y^3/3] |_0^6`

`V = pi[4y - (2y^2)/3+ y^3/27] |_0^6`

Apply definite integration formula:` int_a^b f(y) dy= F(b)-F(a)` .

`V = pi[4(6) - (2(6)^2)/3+(6)^3/27] -pi[4(0) - (2(0)^2)/3+ 0^3/27] `

`V =pi[24 - 24+8] -pi[0 - 0+ 0] `

`V = 8pi` or `25.133` (approximated value)