Calculus of a Single Variable, Chapter 7, 7.2, Section 7.2, Problem 6

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)