Calculus of a Single Variable, Chapter 7, 7.2, Section 7.2, Problem 21

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)