Finding Volume Using the Disk Method

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  • #1
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Hi all, first time here. Huzzah! Looking for help setting up the integral for this:

1. Find the volume of the solid generated by revolving the region bounded by y = x^2 and y = 4x - x^2 around the line y = 6.

2. V = π ʃ [f(x)]^2 dx

3. I've tried every variation of:

π ʃ [(x^2 - 6)^2 - (4x - x^2)^2] dx (0, 2)

and

π ʃ [(x^2 - 6)^2 - (4x - x^2 - 6)^2] dx (0, 2)

I can think of.
 

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  • #2
HallsofIvy
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The second is correct. Since you are rotating around the line y= 6, 6- f(x) is the 'radius' of rotation. [itex]|x^2- 6|[/itex] is the distance from y= 6 to [itex]y= x^2[/itex] so the area of the disk is [itex]\pi(x^2-6)^2[/itex]. [itex]|4x- x^2- 6|[/itex] is the distance from y= 6 to [itex]y= 4x- x^2[/itex] so the area of the disk is [itex]\pi(4x-x^2-6)^2[/itex]. The area of the "washer" between them is [itex]\pi((x^2- 6)^2- (4x-x^2-6)^2[/itex] so the whole volume is
[tex]\pi\int [(x^2- 6)^2- (4x-x^2- 6)^2]dx[/tex]

Now, you say you "tried" that. Exactly what did you do?
 
  • #3
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Now, you say you "tried" that. Exactly what did you do?
I did the algebra:

[tex](x^2 - 6)^2 = x^4 - 12x^2 + 36[/tex]
[tex](4x-x^2 - 6)^2 = x^4 - 8x^3 + 28x^2 - 48x + 36[/tex]
[tex](x^4 - 12x^2 + 36) - (x^4 - 8x^3 + 28x^2 - 48x + 36) = 8x^3 - 40x^2 - 48x[/tex]

Then I integrated:

[tex]\pi\int (8x^3 - 40x^2 - 48x) dx = \pi(2x^4 - (40/3)x^3 - 24x^2)[/tex]

Then I plugged the interval (0, 2) into it:

[tex]\pi(2(2)^4 - (40/3)(2)^3 - 24(2)^2) = \pi(64/3)[/tex]

...which is exactly what the textbook says. *sigh* I feel very foolish; I must have made an algebra mistake during the last step. Thanks anyway!
 
  • #4
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π ʃ [(x^2 - 6)^2 - (4x - x^2 - 6)^2] dx (0, 2)

As a picky point that makes no difference in this problem (and which HallsofIvey pointed out), the 6 and the functions above should be reversed, which can be seen if you draw the pictures and look at the radius of the inner and the outer.

π ʃ [(6-x^2)^2 - (6-4x + x^2)^2] dx

Since they are squared here, it makes no difference, but it would make a difference when you are using the method of cylindrical shells.
 

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