Let C be the curve $\displaystyle {y}={x}^{{3}}+\frac{{1}}{{{12}{x}}}$ for $\displaystyle {0.2}\le{x}\le{0.7}$.

Find the surface area of revolution about the x-axis of C.

First find and simplify $\displaystyle \sqrt{{{1}+{\left({y}'\right)}^{{2}}}}$ = Preview Question 6 Part 1 of 2          (Your answer should not contain a square root.)
Then surface area = Preview Question 6 Part 2 of 2  

Round your area calculation to three decimal places.