Calculus of a Single Variable, Chapter 7, 7.2, Section 7.2, Problem 29

We can use a rectangular strips to represent the region bounded by y=x^2+1 ,y=-x^2+2x+5 , x=0 , and x=3 revolved about the x-axis. As shown on the attached graph, we consider two sets of rectangular strip perpendicular to the x-axis (axis of revolution) to be able to use the Disk Method. This is the case since the upper bound of the rectangular strip differs before and after x=2 .
In this method, we follow the formula: V = int_a^b A(x) dx since we are using a vertical orientation of each rectangular strip with a thickness =dx .
Note: A = pir^2 where r= length of the rectangular strip.
We may apply r = y_(above) - y_(below) .
For the region within the boundary values of x: [ 0,2] , we follow r = (x^2+1)-0=x^2+1
For the region within the boundary values of x: [ 2,3] , we follow r = (-x^2+2x+5)-0 =-x^2+2x+5
Then the integral set-up will be:
V = int_0^2 pi*(x^2+1)^2dx+int_2^4 pi*(-x^2+2x+5)^2dx
Expand: (x^2+1)^2 =(x^2+1)(x^2+1)=x^4+2x^2+1
and (-x^2+2x+5)^2=(-x^2+2x+5)(-x^2+2x+5)=
x^4 - 4x^3 - 6x^2 + 20x + 25
The integral becomes:
V = int_0^2 pi*(x^4+2x^2+1) dx+int_2^3 pi*(x^4 - 4x^3 - 6x^2 + 20x + 25)dx
To evaluate each integrals, we may integrate each term separately using basic integration property: int c f(x) dx = c int f(x) dx and Power rule for integration: int x^n dx = x^(n+1)/(n+1) .
For the first integral:
int_0^2 pi*(x^4+2x^2+1) dx =pi [int_0^2 (x^4+2x^2+1) dx]
=pi*[x^5/5 +(2x^3)/3+x]|_0^2
Apply definite integration formula: int_a^b f(y) dy= F(b)-F(a) .
pi*[x^5/5 +(2x^3)/3+x]|_0^2=pi*[(2)^5/5 +(2(2)^3)/3+(2)]-pi*[(0)^5/5 +(2(0)^3)/3+(0)]
= pi[32/5+16/3+2] - pi[0+0+0]
= pi[206/15] - pi[0]
=(206pi)/15

For the second integral:
int_2^3 pi*(x^4 - 4x^3 - 6x^2 + 20x + 25)dx= pi[int_2^3 (x^4 - 4x^3 - 6x^2 + 20x + 25)dx]
=pi[x^5/5-4*x^4/4-6*x^3/3+20*x^2/2+25x]|_2^3
=pi[x^5/5-x^4-2x^3+10x^2+25x]|_2^3
Apply definite integration formula: int_a^b f(y) dy= F(b)-F(a) .
pi[x^5/5-x^4-2x^3+10x^2+25x]|_2^3
=pi[(3)^5/5-(3)^4-2(3)^3+10(3)^2+25(3)]-pi[(2)^5/5-(2)^4-2(2)^3+10(2)^2+25(2)]
=pi[243/5-81-54+90+75] - [ 32/5-16-16+40+50]
=(393pi)/5 - (322pi)/5
=(71pi)/5
Combing the two definite integral, we get the volume of the solid as:
V =(206 pi)/15+(71pi)/5
V=(419pi)/15 or 87.76 (approximated value.