Calculating the Volume of a Rotated Solid

[endeavor]

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis:
x² - y² = a², x = a + h (where a > 0, h > 0); about the y-axis.

I found the area in terms of y:
A(y) = \pi(2ah + h^2 - y^2)

and the line x = a + h intersects hyperbola at:
y = \pm\sqrt{2ah + h^2}

Thus, the volume is:
V = 2\pi \int^{\sqrt{2ah + h^2}}_{0} (2ah + h^2 - y^2) dy
I simplify this to
V = 2\pi (2ah + h^2)^{3/2}\frac{4}{3} = \frac{8}{3}\pi (2ah + h^2)^{3/2}
however, the answer is not 8/3 pi (2ah + h^2)^(3/2), but 4/3 pi (2ah + h^2)^(3/2). I'm not sure why I have that extra factor of 2 there... Originally, I factored out a 2 so that I would integrate from 0 to \sqrt{2ah + h^2}, instead of -\sqrt{2ah + h^2} to \sqrt{2ah + h^2}.

 

[StatusX]

It looks like you made a mistake evaluating the integral. That should be a 2/3, not 4/3.

 

[endeavor]

It looks like you made a mistake evaluating the integral. That should be a 2/3, not 4/3.

hmm.. you're right. I thought I checked my solution over .. but I missed a sign change from - to +. thanks!