# Cylindrical Rotation Volume Help

1. Jan 2, 2008

1. The problem statement, all variables and given/known data
The region R enclosed by the coordinate axes and the graph of y = k(x-2)^2 is shown above. When this region is revolved around the y - axis, the solid formed has a volume of 8*pi cubic units. What is the value of k?

A) 1
B) 4/3
C) 3/pi
D) 2
E) 3

2. Relevant equations

V = 2*pi f a -> b x [ f(x) - g(x) ] dx

3. The attempt at a solution

Ok so I pluged in 8 pi in for volume and the function for f(x) and o for g(x) and tried solving it.
After simplifying I got 4 = 8/3k
and for k I got 12/8

But that is not an answer choice.

Can I get some help, I would really appreciate it!
Thank You very much!!

2. Jan 2, 2008

### chickendude

Using the shell method, volume is

$$V = 2\pi\int_a^b x * f(x) dx$$

plugging in y=f(x) and the bounds determined by the x and y intercepts
$$= 2\pi k \int_0^2 x * (x-2)^2 dx$$

$$= 2\pi k \int_0^2 x^3 - 4x^2 + 4x dx$$

$$= 2\pi k \left[ \frac{x^4}{4} - \frac{4x^3}{3} + 2x^2 \right]_0^2$$

$$= 2\pi k \left[ 4 - \frac{32}{3} + 8 \right]$$

$$= \frac{8}{3}\pi k$$

since we know the volume is $$8\pi$$

$$8\pi = \frac{8}{3}\pi k$$

$$k = 3$$
bam

3. Jan 2, 2008