Compute the area of the surface $$x^2+y^2-z^2=2y+2z$$ where $-1\leq z \leq 0$
You can get it in the form $x^2+(y-1)^2=(z+1)^2$
I parametrised it as $r(u,v)=(u\cos v, u\sin v+1, u-1)$.
I know that the surface is like a cut version of a cone. But how do I find the limits of $u$ and $v$ so I can do the double integral to find surface area.
Compute following Jacobi Matrix: $J_{ij} = \frac{\partial x_i}{\partial w_j}$. Here $x_1=x,x_2=y,x_3=z,w_1=u,w_2=v$. Defining the $2 \times 2$-Matrix $g_{ij} = \sum_{k=1}^3 J_{ki}J_{kj} = (J^TJ)_{ij}$ you can compute the following Surface integral:
$S = \int_{u_{min}}^{u_{max}}\int_{v_{min}}^{v_{max}} \sqrt{det(g)}dv du$. Since $v$ is an angle variable it holds $v_{min}=0,v_{max}= 2 \pi$. Moreover $u_{min}=0,u_{max}=1$ because $u=z+1$ (in your $z$-component of $r(u,v)$ was a small mistake; it must be $(r(u,v))_z=u-1$) and $-1 \leq z \leq 0$.