The prompt is to find the surface area of a paraboloid between z = 1 and the x-y plane. The paraboloid is given by $z = x^2+y^2$
$$\iint{\sqrt{1+(\frac{\delta{z}}{\delta{x}})^2 + (\frac{\delta{z}}{\delta{y}})^2}}dA$$
$$\iint{r (\sqrt{1 + 4r^2}})drd\theta$$
I know the limits for $\delta{\theta}$ would be from 0 to $2\pi$, but I'm not sure about the limits for $\delta{r}$ but I don't understand how to use the equation z = 1.
UPDATE: from comments $$\int_0^{2\pi}\int_0^1 r\sqrt{1 + 4r^2} drd\theta$$
We see that the part of the paraboloid that is bounded above by the plane $z = 1$ corresponds to $$z = x^2 + y^2 \leq 1$$ which is the solid unit disk, represented in polar coordinates as: $$ 0 \leq r\leq 1 \qquad \qquad 0 \leq \theta \leq 2\pi $$