Finding Volume Using the Disk Method

[1d20]

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.

 

[HallsofIvy]

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

Now, you say you "tried" that. Exactly what did you do?

 

[1d20]

Now, you say you "tried" that. Exactly what did you do?

I did the algebra:

$$(x^2 - 6)^2 = x^4 - 12x^2 + 36$$
$$(4x-x^2 - 6)^2 = x^4 - 8x^3 + 28x^2 - 48x + 36$$
$$(x^4 - 12x^2 + 36) - (x^4 - 8x^3 + 28x^2 - 48x + 36) = 8x^3 - 40x^2 - 48x$$

Then I integrated:

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

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

$$\pi(2(2)^4 - (40/3)(2)^3 - 24(2)^2) = \pi(64/3)$$

...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!

 

[Maxter]

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.