Suppose $S$ is a graph above the region $\Omega$ of the $(xy)$-plane defined by $x^2+y^2\leq a^2$ for a constant $a$, and we know that $|\frac{\partial z}{\partial x}|\leq 2$ and $|\frac{\partial z}{\partial y}|\leq 2$. What is the area of the surface $S$?
I think that the formula I need to use is:
$$
\iint_\Omega f(x,y,z(x,y))\sqrt{1+(\frac{\partial z}{\partial x})^2+(\frac{\partial z}{\partial y})^2} \,dx\,dy
$$
So I think the area after calculation is $\frac{1}{2}a^2\leq Area(S) \leq 3a^2$
But I may have calculated wrong or maybe I'm missing something... Would anyone be able to help?
$|r_x \times r_y| = \sqrt{1+(\frac{\partial z}{\partial x})^2+(\frac{\partial z}{\partial y})^2} = \sqrt{1 + 2^2 + 2^2} = 3$
Surface Area of $S = \displaystyle \int_{-a}^{a}\int_{-\sqrt{a^2-y^2}}^{\sqrt{a^2-y^2}} 3 \,dx\,dy = 3 \pi a^2$
Or write it in polar coordinates for easier integration
$S = \displaystyle \int_{0}^{2 \pi}\int_{0}^{a} 3 r dr d\theta = 3\pi a^2$