Calculus of a Single Variable, Chapter 7, 7.2, Section 7.2, Problem 30

The region bounded by y=sqrt(x) , y =-x/2+4 , x=0 ,and x=8 revolved about the x-axis is shown on the attached image. We may apply Disk Method wherein we use a rectangular strip representation such that it is perpendicular to the axis of rotation. In this case, we need to sets of rectangular strip since the upper bound of the rectangular strip before and after x=4 differs.
The vertical orientation of the rectangular strip shows the thickness of strip =dx.
That will be the basis to use the formula of the Disc method in a form of:
V = int_a^b A(x) dx where A(x) = pir^2 and r =y_(above)-y_(below) .
The r is radius of the disc which is the same as the length of the rectangular strip.
As shown on the attached file, the r= sqrt(x)-0 = sqrt(x) from the boundary values of x=0 to x=4 .
For the boundary values from x=4 to x=8 , we have r=-x/2+4-0 =-x/2+4 .
Then the integral set-up will be:
V = int_0^4 pi(sqrt(x)) ^2dx+ int_4^8 pi(-x/2+4)^2dx
V = int_0^4 pixdx+ int_4^8 pi(-x/2+4)^2dx
We may apply the basic integration property: int c f(x) dx - c int f(x) dx
V = pi int_0^4x dx+ pi int_4^8 (-x/2+4)^2dx
For the integration of piint_0^4x dx , we apply the Power rule for integration: int x^n dx = x^(n+1)/(n+1) .
pi int_0^4xdx =pi*x^((1+1))/((1+1)) |_0^4
= pi*x^(2)/(2) |_0^4
or (pix^2)/2|_0^4
Using the definite integral formula: int_a^b f(x) dx = F(b) - F(a) , we get:
(pix^2)/2|_0^4=(pi(4)^2)/2-(pi(0)^2)/2
= 8pi - 0
=8pi

For the integration of piint_4^8 (-x/2+4)^2dx ,we apply FOIL method to expand.
(-x/2+4)^2 = (-x/2+4)(-x/2+4) = x^2/4-4x+16 .
The integral becomes:
piint_4^8 (x^2/4-4x+16)dx
Apply basic integration property: int (u+-v+-w) dx = int (u) dx +- int (v) dx+- int (w) dx.
piint_4^8 (-x/2+4)^2dx=pi [int_4^8 (x^2/4) dx -int_4^8(4x)^2dx+int_4^8 16 dx]
Apply Power rule for integration: int x^n dx = x^(n+1)/(n+1) and basic integration property: int c dx = cx .
int_4^8 (x^2/4) dx=1/4 int_4^8 x^2 dx
=1/4x^((2+1))/((2+1))|_4^8
=1/4*x^3/3|_4^8
=x^3/12|_4^8
int_4^8 (4x) dx=4 int_4^8 x dx
=4x^(1+1)/(1+1)|_4^8
=4*x^2/2|_4^8
=2x^2|_4^8
int_4^8 16 dx = 16x|_4^8
Then,
pi [int_4^8 (x^2/4) dx -int_4^8(4x)^2dx+int_4^8 16 dx]=pi[x^3/12-2x^2+16x]|_4^8

Using the definite integral formula: int_a^b f(x) dx = F(b) - F(a) , we get:
piint_4^8 (-x/2+4)^2dx=pi[(8)^3/12-2(8)^2+16(8)]-pi[(4)^3/12-2(4)^2+16(4)]
= pi[128/3-128+128] -pi[16/3-32+64]
=(128pi)/3-(112pi)/3
= (16pi)/3

or use u-subtitution by letting u =-x/2+4 then du =-1/2dx or -2 du =dx
pi int (-x/2+4)^2dx =pi int(u)^2* -2 du
= -2pi* u^3/3
Plug-in u= -x/2 +4 or u=(-x+8)/2 on -2pi* u^3/3 , we get:
pi int (-x/2+4)^2dx =-2pi((-x+8)/2)^3/3|_4^8
=-2pi((-x+8)^3/8) *1/3|_4^8
= ((x-8)^3pi)/12|_4^8
Using the definite integral formula: int_a^b f(x) dx = F(b) - F(a) , we get:
piint_4^8 (-x/2+4)^2dx = ((8-8)^3pi)/12 -((4-8)^3pi)/12
= 0 - (-16pi)/3
=(16pi)/3

Then combining the result of the integrals, we get:
V = pi int_0^4xdx+ pi int_4^8 (-x/2+4)^2dx
V=8 pi +(16pi)/3
V =(24pi)/3+(16pi)/3
V=(40pi)/3 or 41.89 (approximated value)