<s>[INST] Surface area of $z=4-x^2-y^2$ over a square region 
Find the surface area of the paraboloid $z=4-x^2-y^2$ over the square region $-2\le x\le 2 $ and $-2\le y\le 2$.

I can parametrize this surface with $x=u,\ y=v,\text{and}  \>z=4-u^2-v^2$,
where $-2\le u\le 2$ and $-2\le v\le 2$. Then I can parametrize the surface with a vector function as follows:
$$\mathbf r(u,v)=\langle u, v ,4-u^2-v^2\rangle $$
Then
$\mathbf r_u(u,v)=\langle 1,0,-2u\rangle,\>
\mathbf r_v(u,v)=\langle 0,1,-2v\rangle$ and
$$\mathbf r_u\times\mathbf r_v=\left|\matrix{\mathbf i&\mathbf j&\mathbf k\cr 1 & 0 & -2u\cr 0 & 1 & -2v}\right|=\langle 2u,2v,1\rangle$$
so $\|\mathbf r_u\times\mathbf r_v\|=\sqrt{4u^2+4v^2+1}$.
The surface area is then defined by
$$A=\int\int_D\|\mathbf r_u\times \mathbf r_v\|\,dA=\int_{-2}^2\int_{-2}^2\sqrt{4u^2+4v^2+1}\,dv\,du$$
But now, how do I perform this integral? Here's an image of this surface.

Update
Due to symmetry, we can calculate the surface area over the region $R=\{(u,v):\ 0\le u\le 2\text{ and }0\le v\le 2\}$, then multiply the result by 4. That is, the surface area is
$$A=4\int_{0}^2\int_{0}^2\sqrt{4u^2+4v^2+1}\,dv\,du$$
Now, it was suggested that I try polar coordinates. Here's an image of the region $R=\{(u,v):\ 0\le u\le 2\text{ and }0\le v\le 2\}$.

Note that the angle of the dashed segment from $(0,0)$ to $(2,2)$ has an angle of $\pi/4$ with the $u$-axis. Also, note that
\begin{align*}
\cos\theta&=\frac2r\\
r\cos\theta&=2\\
r&=\frac{2}{\cos\theta}\\
r&=2\sec\theta
\end{align*}
Again, because of the symmetry of our image, we can determine the surface area over the region $\{(r,\theta):\ 0\le \theta\le \pi/4\text{ and } 0\le r\le 2\sec\theta\}$ and multiply the result by 8. Thus, using
$$ u=r\cos\theta\qquad\text{and}\qquad v=r\sin\theta$$
the surface area is
\begin{align*}
A&=8\int_0^{\pi/4}\int_0^{2\sec\theta}\sqrt{4r^2\cos^2\theta+4r^2\sin^2\theta+1}\ r\,dr\,d\theta\\
A&=8\int_0^{\pi/4}\int_0^{2\sec\theta}\sqrt{4r^2+1}\ r\,dr\,d\theta\\
\end{align*}
Then I was able to integrate and manipulate a bit:
\begin{align*}
A&=8\int_0^{\pi/4}\left[\frac1{12}(4r^2+1)^{3/2}\right]_0^{2\sec\theta}\,d\theta\\
A&=\frac23\int_0^{\pi/4}\left[(16\sec^2\theta+1)^{3/2}-1\right]\,d\theta\\
A&=\frac23\int_0^{\pi/4}(16\sec^2\theta+1)^{3/2}\,d\theta-\frac23\int_0^{\pi/4}d\theta\\
A&=\frac23\int_0^{\pi/4}(16\sec^2\theta+1)^{3/2}\,d\theta-\frac{\pi}{6}
\end{align*}
But now I am unable to calculate this current integral.
Thanks.
Update
Click the link below to see the work I had to do to understand Quanto's nice suggestion.
Surface Area Solution
 [/INST] Let $\sinh t= {\frac{4}{\sqrt{17}}}\tan\theta$ to evaluate
\begin{align}
I &= \int_0^{\pi/4}(16\sec^2\theta+1)^{3/2}\,d\theta\\
&=\frac{17^2}4\int_0^{\sinh^{-1}\frac{4}{\sqrt{17}}}
\frac{\cosh^4 t}{1+\frac{17}{16}\sinh^2t }dt
=34\int_0^{\sinh^{-1}\frac{4}{\sqrt{17}}}
\frac{(1+\cosh 2t)^2}{\frac{15}{17}+\cosh2t}dt\\
 &=\int_0^{\sinh^{-1}\frac{4}{\sqrt{17}}}
\left(34 \cosh2t + 38 + \frac{8}{{15}+17\cosh2t}\right)dt\\
&= 8\sqrt{33} + 38 \sinh^{-1}\frac{4}{\sqrt{17}}+\ 4\int_0^{\sinh^{-1}\frac{4}{\sqrt{17}}}\frac{d(\tanh t)}{16+\tanh^2t}\\
 &= 8\sqrt{33} + 38 \sinh^{-1}\frac{4}{\sqrt{17}}+\cot^{-1}\sqrt{33}
\end{align}
Thus, the surface area is
$$A=\frac23I-\frac{\pi}{6}=\frac23\left( 
 8\sqrt{33} + 38 \sinh^{-1}\frac{4}{\sqrt{17}}+\cot^{-1}\sqrt{33} -\frac\pi4\right)
$$
</s>