I have
$$\sqrt{1+\bigg(\frac{\partial z}{\partial x}\bigg)^2+\bigg(\frac{\partial z}{\partial y}\bigg)^2}=\frac{6}{\sqrt{36-x^2-y^2}}$$
where $z=\sqrt{36-x^2-y^2}$
In the polar form I have to solve
$$\frac{\sigma}{2}= \int_0^\pi \int_0^{6\sin \theta} \frac{6}{\sqrt{36-\rho^2}}\rho\,d\rho\,d\theta=\int_0^\pi\bigg[-6\sqrt{36-\rho^2}\bigg]_0^{6\sin \theta}\,d\theta=\bigg[36\theta -36\sin \theta\bigg]_0^\pi=^?36\pi$$
If I solve this way
$$\frac{\sigma}{4}= \int_0^{\frac{\pi}{2}} \int_0^{6\sin \theta} \frac{6}{\sqrt{36-\rho^2}}\rho\,d\rho\,d\theta=18\pi-36$$
It gives me the real answer $\sigma=72\pi-144$
Why this happens?
You simplified a square root incorrectly. When you substitute the limits $\rho=0$ and $\rho=6\sin\theta$ to the function $\sqrt{36-\rho^2}$ at the upper limit you get $$ \sqrt{36-36\sin^2\theta}=\sqrt{36\cos^2\theta}=6|\cos \theta|. $$
It seems to me that you forgot to take the absolute value. Because $\cos\theta$ is negative in the interval $\theta\in(\pi/2,\pi]$ this makes a difference.
With the absolute value in place you are more or less forced to do the subintervals $\theta\in[0,\pi/2]$ and $\theta\in[\pi/2,\pi]$ separately leading to your other solution.