For the region bounded by `x=y^2` and `x=4 ` revolved about the line
`x=5` , we may apply **Washer method** for the integral
application for the volume of a solid.

The formula for the Washer Method is:

`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`

where *f as function of the outer radius *

*g as a function of the inner
radius*

To determine which form we use, we consider the horizontal rectangular strip representation that is perpendicular to the axis of rotation as shown on the attached image. The given strip have a thickness of "`dy` " which is our clue to use the formula:

`V = pi int_a^b [(f(y))^2-(g(y))^2]dy`

For each radius, we follow the `x_2-x_1` . We have `x_2=5` on both radius since it is a distance between the axis of rotation and each boundary graph.

For the inner radius, we have: `g(y) = 5-4=1` .

For the outer radius, we have: `f(y) = 5-y^2` .

Then the **boundary values of y** are `a=-2` and `b =2`
.

Then the integral will be:

`V = pi int_(-2)^2 [(5-y^2)^2-(1)^2]dy`

Exapnd using the FOIL method on:

`(5-y^2)^2 = (5-y^2)(5-y^2)= 25-10y^2+y^4` and `1^2=1`

The integral becomes:

`V = pi int_(-2)^2 [25-10y^2+y^4-1]dy`

Simplify: `V = pi int_(-2)^2 [24-10y^2+y^4]dy`

Apply basic integration property:

`int (u+-v+-w)dy = int (u)dy+-int (u)dy+-int (v)dy+-int (w)dy`

`V = pi [int_(-2)^2 (24)dy -int_(-2)^2 (10y^2)dy+int_(-2)^2 (y^4) dy]`

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

`V = pi [24y -10y^3/3+ y^5/5]|_(-2)^2`

Apply the definite integral formula:` int _a^b f(x) dx = F(b) - F(a)` .

`V = pi [24(2) -10(2)^3/3+ (2)^5/5] -pi [24(-2) -10(-2)^3/3+ (-2)^5/5]`

`V = pi [48 -80/3+ 32/5] -pi [-48 -(-80)/3+ (-32)/5]`

`V = pi [48 -80/3+ 32/5 +48 -80/3+32/5]`

`V = pi [96 -160/3+ 64/5]`

`V = pi [1440/15 -800/13+ 192/15]`

`V = 832/15pi`

or `V =174.25 ` (approximated value)

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