Calculus of a Single Variable, Chapter 7, 7.2, Section 7.2, Problem 2

To find the volume of a solid by revolving the graph of y =4-x^2 about the x-axis, we consider the bounded region in between the graph and the x-axis. To evaluate this, we apply Disk method by using a rectangular strip perpendicular to the axis of rotation. As shown on the attached image, we consider a vertical rectangular strip with a thickness =dx.
We follow the formula for the Disk Method in a form of: V = int_a^b pir^2 dx or V = pi int_a^b r^2 dx
where r is the length of the rectangular strip.
In this problem, we let the length of the rectangular strip=y_(above)-y_(below) .
Then r = (4-x^2) - 0 = 4-x^2
Boundary values of x: a= -2 to b=2 .
Plug-in the values on the formula V = pi int_a^b r^2 dx , we get:
V =pi int_(-2)^2 (4-x^2)^2 dx
Expand using FOIL method:(4-x^2)^2 = (4-x^2)(4-x^2)= 16-8x^2+x^4 .
The integral becomes:
V =pi int_(-2)^2 (16-8x^2+x^4) 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_(-2)^2(16) dx -int_(-2)^2(8x^2)dx+int_(-2)^2(x^4)dx]
V = pi[16x-(8x^3)/3+x^5/5]|_(-2)^2
Apply definite integration formula: int_a^b f(y) dy= F(b)-F(a) .
V = pi[16(2)-(8(2)^3)/3+(2)^5/5]-pi[16(-2)-(8(-2)^3)/3+(-2)^5/5]
V =pi[32-64/3+32/5]-pi[-32-(-64)/3+(-32)/5]
V =pi[32-64/3+32/5]-pi[-32+64/3-32/5]
V=(256pi)/15 -(-256pi)/15
V=(256pi)/15 +(256pi)/15
V=(512pi)/15 or 107.23 (approximated value)