Cylindrical Rotation Volume Help

[AdiV]

Homework Statement

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

Homework Equations

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

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!

 

[chickendude]

Using the shell method, volume is

[tex]V = 2\pi\int_a^b x * f(x) dx[/tex]

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

[tex]= 2\pi k \int_0^2 x^3 - 4x^2 + 4x dx [/tex]

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

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

[tex]= \frac{8}{3}\pi k[/tex]

since we know the volume is [tex]8\pi[/tex]

[tex]8\pi = \frac{8}{3}\pi k[/tex]

[tex]k = 3[/tex]
bam

 

[AdiV]

Ohh, thanks, yeah I made a calculation error, Thanks!