Find the area of the surface obtained by rotating the curve y= 1+3 x^2 from x=0 to x = 9 about the y-axis.

$We know that if x=f\left(y\right) , a\le y\le b, then the surface area of the solid of revolution about y axis is -$

                                                          $s= {\int }_{y=a}^{b}2\mathrm{\pi x}·\mathrm{ds}$

                                                            $= {\int }_{y=a}^{b}2\mathrm{\pi x}·\left(\frac{\mathrm{ds}}{\mathrm{dy}}\right)·\mathrm{dy}$

                                                             $= {\int }_{y=a}^{b}2\mathrm{\pi x}\sqrt{1+{\left(\frac{\mathrm{dx}}{\mathrm{dy}}\right)}^2}·\mathrm{dy} [ \mathrm{since} {\left(\mathrm{ds}\right)}^2={\left(\mathrm{dx}\right)}^2+{\left(\mathrm{dy}\right)}^2 ]$