Find area of a curved surface within a condition 
Curved surface r given by $\mathbf{r}=(u+v,u-v,uv)^T$. Calculate
  area of a part of the curved surface that satisfies $u^2 + v^2 \leq 1$

Here what I have done:
$\mathbf{r}_u=\frac{\partial \mathbf{r}}{\partial u}= (1, 1, v)^T$
$\mathbf{r}_v=\frac{\partial \mathbf{r}}{\partial u}= (1, 1, v)^T$
Hence
$\mathbf{r}_u \times\ \mathbf{r}_v = (u + v, -u + v, -2)^T$
And
$|\mathbf{r}_u \times\ \mathbf{r}_v| = \sqrt{(2(u^2 + v^2 + 2)}$
Then the area should be
$$S = \iint_D|\mathbf{r}_u \times\ \mathbf{r}_v|dudv=\iint_D\sqrt{(2(u^2 + v^2 + 2)}dudv$$
Let $u = r\cos\theta$, $v = r\sin\theta$ then $r^2 = u^2 + v^2$,
$u^2 + v^2 \leq 1$ give $-1\leq r\leq 1$.
Thus
$$S =\iint_D\sqrt{(2(u^2 + v^2 + 2)}dudv = \int_0^{2\pi}d\theta\int_{-1}^1\sqrt{(2(r^2 + 2)}rdr=0 (!?)$$
Anything wrong with my calculation?

The limits for $r$ should be from 0 to 1 rather than from $-1$ to 1.
$$S=\int_0^1\int_0^{2\pi}r\sqrt{2(r^2+2)}\,d\theta\,dr$$
$$=\int_0^12\pi r\sqrt{2(r^2+2)}\,dr$$
$$=\frac\pi2\int_0^14r\sqrt{2(r^2+2)}\,dr$$
$$=\frac\pi2\left[\frac23(2(r^2+2))^{3/2}\right]_0^1=\pi\left(2\sqrt6-\frac83\right)=7.013\dots$$
