surface area ( double integral)

• August 14th 2007, 02:56 PM
kittycat
surface area ( double integral)
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.
• August 14th 2007, 03:50 PM
galactus
The portion of the cylinder which lies above the xy plane is $z=\sqrt{9-y^{2}}$.

$z_{x}=0, \;\ z_{y}=\frac{-y}{\sqrt{9-y^{2}}}, \;\ z_{x}^{2}+z_{y}^{2}+1=\frac{9}{9-y^{2}}$

$S=\int_{0}^{2}\int_{-3}^{3}\frac{3}{\sqrt{9-y^{2}}}dydx=\int_{0}^{2}\,3{\pi}dx$