Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis:(adsbygoogle = window.adsbygoogle || []).push({});

x^{2}- y^{2}= a^{2}, x = a + h (where a > 0, h > 0); about the y-axis.

I found the area in terms of y:

[tex]A(y) = \pi(2ah + h^2 - y^2)[/tex]

and the line x = a + h intersects hyperbola at:

[tex] y = \pm\sqrt{2ah + h^2} [/tex]

Thus, the volume is:

[tex] V = 2\pi \int^{\sqrt{2ah + h^2}}_{0} (2ah + h^2 - y^2) dy [/tex]

I simplify this to

[tex] V = 2\pi (2ah + h^2)^{3/2}\frac{4}{3} = \frac{8}{3}\pi (2ah + h^2)^{3/2} [/tex]

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 [tex]\sqrt{2ah + h^2}[/tex], instead of [tex]-\sqrt{2ah + h^2}[/tex] to [tex]\sqrt{2ah + h^2}[/tex].

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Find the volume of the solid

**Physics Forums | Science Articles, Homework Help, Discussion**