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

As shown on the attached image, we are using * vertical rectangular strip that is perpendicular to the x-axis (axis of...*

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

As shown on the attached image, we are using ** vertical rectangular strip that is perpendicular to the x-axis (axis of revolution)** with a thickness of `"dx"` . In line with this, we will consider the formula for the Washer Method as:

`V = pi int_a^b [(f(x))^2-(g(x))^2]dx`

where `f(x)` as function of the outer radius, `R`

`g(x)` as a function of the inner radius, `r`

For each radius, we follow the `y_(above) - y_(below)` , we have `y_(below)=0` since it a distance between the axis of rotation and each boundary graph.

For the inner radius, we have: `g(x) =2-0=2`

For the outer radius, we have:` f(x) =(4-x^2/4 )-0=4-x^2/4`

To determine the boundary values of x, we equate the two values of y's:

`4-x^2/4 =2`

`-x^2/4 =2-4`

`-x^2/4 =-2`

`(-4)(-x^2/4 ) =(-4)(-2)`

`x^2=8 ` then ` x= +-sqrt(8) ` or `+2sqrt(2) ` and` -2sqrt(2)`

Then, boundary values of x: `a=-2sqrt(2)` and `b=2sqrt(2)` .

Plug-in the values in the formula `V = pi int_a^b( (f(x))^2 -(g(x))^2) dx` , we get:

`V =pi int_(-2sqrt(2))^(2sqrt(2)) [(4-x^2/4)^2 -2^2]dx` .

Expand using the FOIL method on:` (4-x^2/4)^2 = (4-x^2/4)(4-x^2/4)= 16-2x^2+x^4/16` and `2^2=4` .

The integral becomes:

`V =pi int_(-2sqrt(2))^(2sqrt(2)) [16-2x^2+x^4/16 -4]dx`

`V =pi int_(-2sqrt(2))^(2sqrt(2)) [12-2x^2+x^4/16 ]dx`

Apply basic integration property: `int (u+-v+-w)dx = int (u)dx+-int (v)dx+-int(w)dx` to be able to integrate them separately using Power rule for integration: `int x^n dx = x^(n+1)/(n+1)` .

`V =pi *[int_(-2sqrt(2))^(2sqrt(2))(12) dx -int_(-2sqrt(2))^(2sqrt(2)) (2x^2) dx + int_(-2sqrt(2))^(2sqrt(2)) (x^4/16)dx]`

`V =pi *[12x-2 *x^3/3+1/16*x^5/5 ]|_(-2sqrt(2))^(2sqrt(2))`

`V =pi *[12x-(2x^3)/3+x^5/80 ]|_(-2sqrt(2))^(2sqrt(2))`

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

`V =pi *[12(2sqrt(2))-(2(2sqrt(2))^3)/3+(2sqrt(2))^5/80 ]-pi *[12(-2sqrt(2))-(2(-2sqrt(2))^3)/3+(-2sqrt(2))^5/80 ]`

`V =pi *[24sqrt(2)-(32sqrt(2))/3+(8sqrt(2))/5 ] -pi *[-24sqrt(2)+(32sqrt(2))/3-(8sqrt(2))/5 ]`

`V =(224sqrt(2)pi)/15 -(-224sqrt(2)pi)/15`

`V =(224sqrt(2)pi)/15 +(224sqrt(2)pi)/15`

`V =(448sqrt(2)pi)/15` or `132.69` (approximated value)