I have to find the area of the surface $$x+y+z=5$$ localizated above the region $x^{2}+y^{2}\leq9$$.
I did: $$z=5-x-y\Rightarrow z_{x}=z_{y}=-1$$ So the area is $$\int\int_{R}\sqrt{1+z_{x}^{2}+z_{y}^{2}}dxdy$$ where $R$ is the region $x^{2}+y^{2}\leq9$$
Using polar coordinates, I got: $$x=r\sin\theta, y=r\cos\theta$$ where $$r\in[0,3]$$ and $$\theta\in[0,2\pi]$$
So the area is $$A=\int_{0}^{2\pi}\int_{0}^{3}\sqrt{3}rdrd\theta$$ And when I evaluate it, I get $9\sqrt{3}\pi$, but the answer in the book is $6\pi$. Where is my mistake?
Your answer squares with the one I got using plain analytic geometry. The area of an ellipse is $\pi ab$, where $a$ and $b$ are the semimajor and semiminor axes, that is, the greatest and least distance from the center of the ellipse.
The center of the ellipse is at $(0,0,5)$. Since $z=5-x-y$, the greatest $z$-value occurs when $x+y$ is least, at $x = y = -\frac{3}{\sqrt{2}}$. So the semimajor axis is the distance from $(0,0,5)$ to $\left(-\frac{3}{\sqrt{2}},-\frac{3}{\sqrt{2}},5+2\sqrt{3}\right)$, and this is $3\sqrt{3}$.
In the perpendicular direction, the semiminor axis is the distance from $(0,0,5)$ to $\left(\frac{3}{\sqrt2},-\frac{3}{\sqrt2},5\right)$, which is $3$. So the area is $$ \pi(3)(3\sqrt{3}) = 9\pi\sqrt{3} $$