Find the surface area of the portion of the cylinder y^2 + z^2 = 9 that is above the rectangle R = {(x,y) such that 0< or = x < or = 2 , -3 < or = y<or = 3}
Thank you very much.
The portion of the cylinder which lies above the xy plane is $\displaystyle z=\sqrt{9-y^{2}}$.
$\displaystyle z_{x}=0, \;\ z_{y}=\frac{-y}{\sqrt{9-y^{2}}}, \;\ z_{x}^{2}+z_{y}^{2}+1=\frac{9}{9-y^{2}}$
$\displaystyle S=\int_{0}^{2}\int_{-3}^{3}\frac{3}{\sqrt{9-y^{2}}}dydx=\int_{0}^{2}\,3{\pi}dx$