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...
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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.
