# `y = x , y = 3 , x = 0` Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line y = 4

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For a region bounded by `y=x` (slant line), `x=0` (along the vertical axis), and `y = 3` (horizontal line) revolved about the line `y=4` , we may apply the **Washer Method**. It considers multiple disc with a hole. Basically, it can be just two disc method in which we take the difference of the volume of the bigger and smaller disc.

It follows the formula: `A = pi(R_(outer))^2-(r_(i.n.n.e.r))^2]`

Let `R_(outer)` as a function of `f` and` r_(i.n.n.e.r) ` as function of `g` .

Then `V =pi int_a^b[(f(x))^2-(g(x))^2]dx` or `V =pi int_a^b[(f(y))^2-(g(y))^2]dy` .

We use a rectangular strip representation that is perpendicular to the axis of rotation as shown on the attached image.

For the inner radius, we have:

`g(x)=4-3=1 `

For the outer radius, we have: `f(x)=4-x`

The boundary values of `x` will be `a=0 ` to `b=3` .

The integral to approximate the volume of the solid is:

`V=pi int_0^3 [(4-x)^2-1^2] dx `

Expand using FOIL method on `(4-x)^2 = (4-x)*(4-x) = 16-8x+x^2` and `1^2=1` .

The integral becomes:

`V=pi int_0^3 [16-8x+x^2 -1] dx `

Simplify:`V=pi int_0^3 [15-8x+x^2] dx`

Apply basic integration property: `int (u+-v+-w) dx= int (u) dx+-int (v) dx+-int (w) dx`

`V=pi [ int_0^3 15 dx - int_0^3 8xd +int_0^3 x^2 dx]`

Apply basic integration property:` int c dx = cx` and Power rule for integration: `int x^n dx= x^(n+1)/(n+1).`

`V=pi [15x- 8*x^2/2 + x^3/3]|_0^3`

`V=pi [15x- 4x^2 + x^3/3]|_0^3`

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

`V=pi [15*(3)- 4(3)^2 +(3)^3/3]-pi [15(0)- 4(0)^2 + (0)^3/3]`

`V=pi [45- 36+9] -pi [0-0+0]`

`V = 18pi` or `56.55` (approximated value)