# Volume between hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and line $x = 2a$ around $y$ axis

I'm trying to calculate the volume between the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and the line $x = 2a$ around the $y$ axis using two methods but I'm getting different answers:

1. Using volume of a solid of revolution with circular-ring method (the usual formula is multiplied by $2$ because half the interval is defined for the integral): $$\begin{eqnarray} V &=& 2 \pi \int_0^{b\sqrt{3}} [(2a)^2 - (\frac{a}{b}\sqrt{y^2+b^2})^2] \, \textrm{d}y \\ &=& \frac{2 \pi a^2}{b^2} \left[3 b^2 y - \frac{y^3}{3}\right]_0^{b \sqrt{3}} \\ &=& 4 \sqrt{3} \pi a^2 b \\ \end{eqnarray}$$

2. Using volume of a solid of revolution with cylindrical-shell method: $$\begin{eqnarray} V &=& 2 \pi \int_a^{2a} [x (\frac{2b}{a} \sqrt{x^2 - a^2})] \, \textrm{d}x \\ &=& 4 \pi a b \left[ \frac{(x^2 - a^2)^{3/2}}{3 a^3} \right]_a^{2a} \\ &=& 4 \sqrt{3} \pi a b \\ \end{eqnarray}$$

The second answer is the one from my book but I'd really like to understand why the first answer is different.

• Thanks, so my book was wrong once again. By the way, what technique did you use to integrate $x \sqrt{x^2 - a^2}$? Jan 11 '16 at 3:08
• Just substitution, essentially, after recognizing that $\frac{d}{dx}(x^2-a^2)^{3/2} = 3x$. I suppose the "calculus book" way of doing it is to use the substitution $x = a\sec u$ and the identity $\sec^2 u - 1 = \tan^2 u$. What book is this? Jan 11 '16 at 3:12
• Thanks again, $x = a \sec \theta$ worked. My book is an old edition of The Calculus with Analytic Geometry by Louis Leithold. Jan 11 '16 at 3:16