Answer
(a) i) $y = \tan x$ then $dy/dx = \sec^{2} x$ and $ds = \sqrt{1+(dy/dx)^{2}}dx = \sqrt{1+\sec^{4} x}dx$
$S = \int 2\pi y ds = \int ^{\frac{\pi}{3}}_{0} 2\pi \tan x \sqrt{1+\sec^{4}x}dx$
ii) $S = \int 2\pi x ds = \int^{\frac{\pi}{3}}_{0}2\pi x \sqrt{1+\sec^{4} x} dx$
(b) i) 10.5017 ii) 7.9353
Work Step by Step
(a) i) $y = \tan x$ then $dy/dx = \sec^{2} x$ and $ds = \sqrt{1+(dy/dx)^{2}}dx = \sqrt{1+\sec^{4} x}dx$
$S = \int 2\pi y ds = \int ^{\frac{\pi}{3}}_{0} 2\pi \tan x \sqrt{1+\sec^{4}x}dx$
ii) $S = \int 2\pi x ds = \int^{\frac{\pi}{3}}_{0}2\pi x \sqrt{1+\sec^{4} x} dx$
(b) i) 10.5017 ii) 7.9353
