Finding Area Underneath Variable Curve

[Vile Smile]

Homework Statement

The region enclosed by [PLAIN]http://latex.codecogs.com/gif.latex?y^2%20=%20kx and the line [URL]http://latex.codecogs.com/gif.latex?x%20=%20\frac{1}{4}k[/URL] is revolved about the line [URL]http://latex.codecogs.com/gif.latex?x%20=%20\frac{1}{2}k.[/URL] Use cylindrical shells to find the volume of the resulting solid. (Assume k > 0)

Homework Equations

[URL]http://latex.codecogs.com/gif.latex?V%20=%202\pi\int_{a}^{b}xf(x)[/URL]

The Attempt at a Solution

I started by trying to find the interval. I've managed to come up [URL]http://latex.codecogs.com/gif.latex?b=\frac{1}{4}k,[/URL] but I'm unsure of how to find a. In addition, because the area is being revolved around a vertical axis, I changed [PLAIN]http://latex.codecogs.com/gif.latex?y^2%20=%20kx to [URL]http://latex.codecogs.com/gif.latex?y=\sqrt{kx}.[/URL]

Here is my current equation:

[URL]http://latex.codecogs.com/gif.latex?V%20=%202\pi\int_{a}^{\frac{1}{4}k}(\frac{1}{2}k-x)(\sqrt{kx})[/URL]I need to find a. Also, I would appreciate if someone told me if I've taken the right steps so far in solving the problem.

 

[JThompson]

The domain of $$y^{2}=kx$$ is $$[0 , \infty]$$, so x=0 would have to be your lower limit.

 

[JThompson]

Your problem also looks correct so far, though you may consider whether they are asking about $$y=\sqrt{kx}$$ or $$y^{2}=kx$$. The region between $$y^{2}=kx$$ and $$x=\frac{1}{4}k$$ is larger than the region used in your integral.