Chapter 8 - Further Applications of Integration - 8.2 Area of a Surface of Revolution - 8.2 Exercises - Page 595: 13

$\frac{21\pi}{2}$

$x = \frac{1}{3} (y^{2}+2)^{3/2}$ then $dx/dy = \frac{1}{2} (y^{2} + 2)^{1/2} (2y) = y \sqrt{y^{2} +2}$ and $1+(dx/dy)^{2} = 1+y^{2} (y^{2}+2) = (y^{2}+1)^{2}$ So $S = 2\pi \int^{2}_{1} y(y^{2}+1) dy = 2\pi [\frac{1}{4} y^{4} +\frac{1}{2} y^{2}]^{2}_{1} = 2\pi (4+2-\frac{1}{4} -\frac{1}{2}) = \frac{21\pi}{2}$