Finding Volume Using the Disk Method

1d20
Messages
12
Reaction score
0
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.
 
Physics news on Phys.org
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?
 
HallsofIvy said:
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!
 
1d20 said:
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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top