Three Surface Integrals Could someone assist with the following three surface integrals?   
Q1 The portion of the cone $z=\sqrt{x^2+y^2}$ that lies inside the cylinder $x^2+y^2 =2x$.   
Q2 The portion of the paraboloid $z=1-x^2-y^2$ that lies above the $xy$-plane.
Q3 The portion of the paraboloid $2z = x^2+y^2$ that is inside the cylinder $x^2+y^2=8$.
Any assistance will be greatly appreciated.

You have $z=f(x,y)$ and the surface area is $$\int\int_A\sqrt{1+f_x^2+f_y^2}dxdy$$ where $A$ is the projection of the surface area on the $xy$-plane.
Use polar coordinates and so $dA=rdrd\theta$  


*

*$r=0\to r=2\cos\theta,\theta=-\pi/2\to\pi/2$

*$r=0\to r=1,\theta=0\to2\pi$

*$r=0\to r=\sqrt8,\theta=0\to2\pi$
For the first one, $z=\sqrt{x^2+y^2}$ and so
$$\int\int_A\sqrt{1+\frac{x^2}{x^2+y^2}+\frac{y^2}{x^2+y^2}}dxdy
\\=\int_{-\pi/2}^{\pi/2}\int_0^{2\cos \theta}\sqrt{2}rdrd\theta
\\=\frac{\sqrt{2}}{2}\int_{-\pi/2}^{\pi/2}r^2|_0^{2\cos \theta}d\theta$$   



