What is wrong with this approach? I am given a problem to find the surface area of the cylinder $z^2+y^2=9$ above the rectangle defined by the points $(0,0),(4,0),(0,2),(4,2)$.
Instead of trying to integrate on with any respect to $z$, I tried to treat the domain purely a matter of $x$ and $y$.
I start by integrating with respect to $z_y$.  I isolate $z$ for $z=\sqrt{9-y^2}$ (we can omit the $+-$ because we are only interested in the area above the rectangle.)  
I get $z_y = \frac{-2y}{\sqrt{9-x^2}}$, this reduced to a terrifying $\displaystyle \iint \limits_R {\sqrt{\frac{-2y}{\sqrt{9-y^2}}+1}}$ for $R=[0,2]\mathrm{x}[0,4]$.  Not likely to happen...
Now, I try to integrate by projecting a polar coordinate system onto the $Y\mathrm{-}Z$ plane through which we integrate.  Allowing $z$ to be $y$ and $y$ to be $x$.... ah, 

simpler still....

1.) Take the circumference of a circle which becomes our cylinder extruded indefinitely along the x axis, above the z axis, but divide it in half since we only want the top.  This $6\pi$.
2.) Realize the region of $z$ projected onto the $X-Y$ plane is different from the $R$ specified in the problem... values for the $y$-axis only extend out to three, whereas the $x$-axis extends indefinitely.  Therefore, we need only concern ourselves with the $x$-axis.
3.) Multiply the range of the x axis by the circumference, which happens to be $2\cdot6\pi=12\pi$.
Which is wrong.  The answer is $12\arcsin(2/3)$, not $12\pi$.
So confused... why did my simple approach just above not work!?  How are you supposed to get the answer?
I really appreciate any help!
 A: I guess your biggest mistake is not using the area formula correctly.
$z=\sqrt{9-y^2}$. So the area is
$$
\int_0^2\int_0^4\sqrt{1+z_x^2+z_y^2}\,dxdy=\int_0^2\int_0^4\sqrt{1+\frac{y^2}{{9-y^2}}}\,dxdy=\int_0^2\int_0^4\sqrt{\frac{9}{{9-y^2}}}\,dxdy\\=3\int_0^2\int_0^4({{9-y^2}})^{-1/2}\,dxdy=12\int_0^2({{9-y^2}})^{-1/2}\,dy=12\,\arcsin\left.\left(\frac{y}3\right)\right|_0^2=12\,\arcsin\left(\frac23\right)
$$
A: I suspect you haven't handled the "projection factor" of the area of the cylinder onto the xy-plane correctly.  But you have a good idea of what you want to look at.  Consider using cylindrical coordinates with the "z-axis" along the axis of the cylinder.  The "height" then is 4 units, as you've said.  You want the portion of the circumference of the circle of radius 3 represented by the cross-section from angle 0  to the angle represented by going 2 units "horizontally" away from the "vertical" on this circle in the yz-plane.  (That angle does not come out evenly, but it does have a simple description from trigonometry.)
