Let C be the curve $\displaystyle {y}={9}\sqrt{{{x}}}$ for $\displaystyle {1.2}\le{x}\le{3.8}$.

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

Surface area = $\displaystyle {\int_{{1.2}}^{{3.8}}}{f{{\left({x}\right)}}}\ {\left.{d}{x}\right.}$ where
f( x ) = Preview Question 6 Part 1 of 2  
Now integrate to find surface area = $\displaystyle {\int_{{1.2}}^{{3.8}}}{f{{\left({x}\right)}}}\ {\left.{d}{x}\right.}$ = Preview Question 6 Part 2 of 2