Calculus of a Single Variable, Chapter 7, 7.2, Section 7.2, Problem 24

The region bounded by y=xsqrt(4-x^2) and y=0 revolved about the x-axis is shown on the attached image.We may apply the Disk method using a rectangular strip perpendicular to the axis of revolution. As shown on the attached image, the thickness of the rectangular strip is "dx" with a vertical orientation perpendicular to the x-axis (axis of revolution).
We follow the formula for the Disk method:V = int_a^b A(x) dx where disk's base area is A= pi r^2 with r =y=f(x) .
Note: r = length of the rectangular strip. We may apply r = y_(above)-y_(below) .
Then r =(xsqrt(4-x^2))- 0 =xsqrt(4-x^2) .
The boundary values of x are a=-2 to b=2 .
Plug-in the f(x) and the boundary values to integral formula, we get:

V = int_(-2)^2 pi(xsqrt(4-x^2))^2 dx
Simplify:
V = int_(-2)^2 pix^2(4-x^2) dx
V = int_(-2)^2 pi*(4x^2-x^4) dx
Apply basic integration property: intc*f(x) dx = c int f(x) dx
V = pi int_(-2)^2 (4x^2-x^4) dx
Apply basic integration property:int (u-v)dx = int (u)dx-int (v)dx .
V = pi *[ int_(-2)^2 (4x^2)dx -int_(-2)^2(x^4) dx]
Apply Power rule for integration: int x^n dx= x^(n+1)/(n+1) .
V = pi *[(4x^3)/3 -x^5/5]|_(-2)^2
Apply definite integration formula: int_a^b f(y) dy= F(b)-F(a) .
V = pi *[(4(2)^3)/3 -(2)^5/5] -pi *[((-2)^3)/3 -(-2)^5/5]
V = pi *[32/3 - 32/5] -pi *[(-32)/3 -(-32)/5]
V = pi *[160/15 - 96/15] -pi *[(-160)/15 -(-96)/15]
V = pi *[64/15 ] -pi *[(-64)/15 ]
V =(64pi)/15 -( -64pi)/15
V =(64pi)/15 +64pi/15
V =(128pi)/15 or 26.81 (approximated value)