$$Z = \log (x^2 + y^2) $$ $$1< x^2 + y^2 < 5 $$ Find the area of the surface.
My attempt: If $z=f(x,y)$, then the normal vector is $$\sqrt { \frac{df}{dx} ^2 + \frac{df}{dy} ^2 + 1} $$ which gives me $$\sqrt { \displaystyle\frac{4}{(x^2 + y^2)} +1}$$
Now I parameterize $x$ and $y$ using polar coordinates:
Let $x = r \cos t$ and $y = r \sin t$, where $1 \le r \le \sqrt{5}$ and $0 \le t \le 2\pi$.
My normal vector would then be $\sqrt{\frac{4}{r^2} +1}$.
Multiplying by the Jacobian for polar coordinates , I get $\sqrt{4+r^2}$.
Then I have to integrate that with respect to $dr$ and $dt$ to get the surface area, is that right??