# `y = x^2 + 1 , y= -x^2 + 2x +5 , x=0 , x=3` Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis.

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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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.

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