Find the area of the surface obtained by rotating the curve about x-axis y= $(9-x^2)^.5$ $; 0<=x<=2$ I did the $\int_0^2$ and put that into the $$SA = \int_0^2 2\pi (\sqrt{9-x^2})*\sqrt{1+(-x/\sqrt{9-x^2})^2} $$ My answer came to be $6\pi$ It is supposed to be $12\pi$ however I manged to cancel my remaining $2$ after multiplying $(f'(x))^2$ letting $u= 9-x^2$ and $du = -2x$ making $dx= (1/-2x)$

\begin{align} \int_0^2 2\pi (\sqrt{9-x^2})\cdot\sqrt{1+(-x/\sqrt{9-x^2})^2}dx &=\int_0^2 2\pi (\sqrt{9-x^2})\cdot\sqrt{\dfrac{9-x^2+x^2}{9-x^2}}dx\\\ &=\int_0^2 2\pi\sqrt{9}\,dx\\\ &=6\pi\int_0^2dx\\\ &=6\pi x{\huge\vert}_0^2=12\pi \end{align}