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: 11

$$S = \frac{{16,911}}{{1024}}\pi $$

$$\eqalign{ & 8x{y^2} = 2{y^6} + 1,\,\,\,\,1 \leqslant y \leqslant 2 \cr & {\text{Solve for }}x \cr & x = \frac{{2{y^6} + 1}}{{8{y^2}}} \cr & x = \frac{1}{4}{y^4} + \frac{1}{8}{y^{ - 2}} \cr & {\text{Calculate }}\frac{{dx}}{{dy}} \cr & \frac{{dx}}{{dy}} = {y^3} - \frac{1}{4}{y^{ - 3}} \cr & {\text{Use }}S = \int_a^b {2\pi x\sqrt {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} } dy \cr & S = \int_1^2 {2\pi \left( {\frac{{2{y^6} + 1}}{{8{y^2}}}} \right)\sqrt {1 + {{\left( {{y^3} - \frac{1}{4}{y^{ - 3}}} \right)}^2}} } dy \cr & S = \frac{1}{4}\pi \int_1^2 {\left( {2{y^4} + {y^{ - 2}}} \right)\sqrt {1 + {{\left( {{y^3} - \frac{1}{4}{y^{ - 3}}} \right)}^2}} } dy \cr & S = \frac{1}{4}\pi \int_1^2 {\left( {2{y^4} + {y^{ - 2}}} \right)\sqrt {1 + {{\left( {{y^3}} \right)}^2} + 2\left( {{y^3}} \right)\left( { - \frac{1}{4}{y^{ - 3}}} \right) + {{\left( { - \frac{1}{4}{y^{ - 3}}} \right)}^2}} } dy \cr & S = \frac{1}{4}\pi \int_1^2 {\left( {2{y^4} + {y^{ - 2}}} \right)\sqrt {1 + {{\left( {{y^3}} \right)}^2} - \frac{1}{2} + {{\left( { - \frac{1}{4}{y^{ - 3}}} \right)}^2}} } dy \cr & S = \frac{1}{4}\pi \int_1^2 {\left( {2{y^4} + {y^{ - 2}}} \right)\sqrt {{{\left( {{y^3} + \frac{1}{4}{y^{ - 3}}} \right)}^2}} } dy \cr & S = \frac{1}{4}\pi \int_1^2 {\left( {2{y^4} + {y^{ - 2}}} \right)\left( {{y^3} + \frac{1}{4}{y^{ - 3}}} \right)} dy \cr & {\text{Distribute and simplify}} \cr & S = \frac{1}{4}\pi \int_1^2 {\left( {2{y^7} + \frac{1}{2}y + y + \frac{1}{4}{y^{ - 5}}} \right)} dy \cr & S = \frac{1}{4}\pi \int_1^2 {\left( {2{y^7} + \frac{3}{2}y + \frac{1}{4}{y^{ - 5}}} \right)} dy \cr & {\text{Integrate}} \cr & S = \frac{1}{4}\pi \left[ {\frac{1}{4}{y^8} + \frac{3}{4}{y^2} - \frac{1}{{16}}{y^{ - 4}}} \right]_1^2 \cr & S = \frac{1}{4}\pi \left[ {\frac{1}{4}{{\left( 2 \right)}^8} + \frac{3}{4}{{\left( 2 \right)}^2} - \frac{1}{{16}}{{\left( 2 \right)}^{ - 4}}} \right] - \frac{1}{4}\pi \left[ {\frac{1}{4} + \frac{3}{4} - \frac{1}{{16}}} \right] \cr & S = \frac{1}{4}\pi \left( {\frac{{17151}}{{256}}} \right) - \frac{1}{4}\pi \left( {\frac{{15}}{{16}}} \right) \cr & S = \frac{{16,911}}{{1024}}\pi \cr} $$