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:
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): $$ \begin{eqnarray} V &=& 2 \pi \int_0^{b\sqrt{3}} [(2a)^2 - (\frac{a}{b}\sqrt{y^2+b^2})^2] \, \textrm{d}y \\ &=& \frac{2 \pi a^2}{b^2} \left[3 b^2 y - \frac{y^3}{3}\right]_0^{b \sqrt{3}} \\ &=& 4 \sqrt{3} \pi a^2 b \\ \end{eqnarray} $$
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.
The second integral is wrong. \begin{equation*} \int \left(x\left(\frac{2b}{a}\sqrt{x^2-a^2}\right)\right)\,dx = \frac{2b}{3a}\int 3x\sqrt{x^2-a^2}\,dx = \frac{2b}{3a}(x^2-a^2)^{3/2}, \end{equation*} so the definite integral is \begin{equation*} V = 2\pi\int_a^{2a} \left(x\left(\frac{2b}{a}\sqrt{x^2-a^2}\right)\right)\,dx = \frac{4b\pi}{3a}(x^2-a^2)^{3/2}\big\lvert_a^{2a} = 4 \sqrt{3} \pi a^2 b . \end{equation*}