Chapter 8 - Further Applications of Integration - 8.2 Area of a Surface of Revolution - 8.2 Exercises - Page 595: 14

$\frac{\pi}{24}(65\sqrt{65} - 17\sqrt{17})$

$x = 1+2y^{2}$ then $1+(dx/dy)^{2} = 1+(4y)^{2} = 1 + 16y^{2}$ So $S = 2\pi \int^{2}_{1} y \sqrt{1+16y^{2}} dy = \frac{\pi}{16} \int^{2}_{1} (16y^{2}+1)^{1/2} 32y dy = \frac{\pi}{24}(65\sqrt{65} - 17\sqrt{17})$