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].

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# Homework Help: Find the volume of the solid

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