Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.5 Area Of A Surface Of Revolution - Exercises Set 5.5 - Page 380: 2

$$S = \frac{{\left( {17\sqrt {17} - 5\sqrt 5 } \right)\pi }}{6}$$

$$\eqalign{ & y = \sqrt x ,\,\,\,\,1 \leqslant x \leqslant 4 \cr & {\text{Calculate }}\frac{{dy}}{{dx}} \cr & \frac{{dy}}{{dx}} = \frac{1}{{2\sqrt x }} \cr & {\text{Use }}S = \int_a^b {2\pi y\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } dx \cr & S = \int_1^4 {2\pi \left( {\sqrt x } \right)\sqrt {1 + {{\left( {\frac{1}{{2\sqrt x }}} \right)}^2}} } dx \cr & S = 2\pi \int_1^4 {\sqrt x \sqrt {1 + \frac{1}{{4x}}} } dx \cr & S = 2\pi \int_1^4 {\frac{{\sqrt x }}{{2\sqrt x }}\sqrt {4x + 1} } dx \cr & S = \pi \int_1^4 {\sqrt {4x + 1} } dx \cr & {\text{Integrate}} \cr & S = \frac{\pi }{4}\left[ {\frac{{2{{\left( {4x + 1} \right)}^{3/2}}}}{3}} \right]_1^4 \cr & S = \frac{\pi }{6}\left[ {{{\left( {4x + 1} \right)}^{3/2}}} \right]_1^4 \cr & S = \frac{\pi }{6}\left[ {{{\left( {16 + 1} \right)}^{3/2}} - {{\left( 5 \right)}^{3/2}}} \right] \cr & S = \frac{{\left( {17\sqrt {17} - 5\sqrt 5 } \right)\pi }}{6} \cr} $$