$z=\sqrt{x^2+y^2}$ is the surface we working on.
I am a bit stuck on choosing the limits for this problem, I have done the following:
$J(\text{jacobian})=\sqrt{Z^2_x+Z^2_y+1}=\left(\dfrac{x^2+y^2}{(\sqrt{x^2+y^2})^2}+1\right)^{\frac{1}{2}}=\sqrt{2}$ $$\iint_S f \, dS=\iint \sqrt{2} \, dz \, d\theta.$$
I can't explain why I chose $dz, d\theta$, I guess since the one surface is bounded by $z$ and $x^2+y^2=x$ looks like the equation of a circle so $\theta$ :)
Anyway, I know that $z \left[0:\sqrt{x^2+y^2} \right] \rightarrow \left[0:r \right]$
and
$x^2+y^2=x\rightarrow r^2=r\cos(\theta)$, I tried to solve for $\theta$ from that but no luck. Please assist -Thanks.
I looked at this post here and it doesn't really address my issue.
No need to solve for $\theta$. Notice that $z$ is not an independent variable, thus not a good choice. Using $x = r \cos(\theta)$ and $y = r \sin(\theta)$ you obtain $z=r$. The length of the normal vector is $$\sqrt{ \left( \frac{\partial z}{\partial r} \right)^2 + \left( \frac{\partial z}{\partial \theta} \right)^2 + 1 } = \sqrt{2}.$$ The surface area is $$\iint_S dS = \iint_S \| {\mathbf r}_u \times {\mathbf r}_v \| \, dA.$$ Notice that this is not a Jacobian. Jacobians arise in changes of variable of the form $\mathbb{R}^2 \to \mathbb{R}^2$ or $\mathbb{R}^3 \to \mathbb{R}^3$. Since this is the graph of a function, the natural parameterization is $${\mathbf r} = (r \cos(\theta), r \sin(\theta), r).$$ The limits for $\theta$ are $0 \leq \theta \leq 2 \pi$ and for $r$ are $0 \leq r \leq \cos(\theta)$. Your area integral becomes $$\iint dS = \int_0^{2 \pi} \int_0^{\cos(\theta)} \sqrt{2} r \, dr \,d \theta.$$