A circular metal wire $S$ in 3-space is described by $S=\{x^2+y^2+z^2=4\} $ a.) A circular metal wire $S$ in 3-space is described by $S=\{x^2+y^2+z^2=4\} $ and has density $\sigma (x,y,z) = \sqrt{x^2+y^2}$ at the point $(x,y,z)$. Find the mass of $S$. 
My attempt: 
$$\int_{0}^{2\pi}\int_{0}^{2} r^2 drd\theta = \frac{16\pi}{3}$$
Is this the correct approach/solution to this problem. Looking for verification please. 
 A: You want to compute the surface integral along $S$ of the function $\sigma (x,y,z) = \sqrt{x^2+y^2}$ wich is:
$$\iint_S \sigma(\Phi(\theta,\phi)) ~||\Phi_\theta(\theta,\phi)\times \Phi_{\phi}(\theta,\phi)|| ~d\theta d\phi$$ where $\Phi(\theta,\phi)$ is a parametrization of $S$. Let $\Phi = (2\cos \theta \sin \phi, 2\sin\theta \sin \phi, 2\cos \phi)$.
$$\iint_S \sigma(\Phi) ~||\Phi_\theta \times \Phi_{\phi}|| ~d\theta d\phi  = \int_0^{\pi} \int_0^{2\pi} 2\sqrt{ \cos^2(\theta) \sin^2(\phi)+\sin^2(\theta) \sin^2(\phi)}~ ~4\sin\phi~~d\theta d\phi = $$
$$= \int_0^{\pi} \int_0^{2\pi} 8 \sin^2(\phi) ~ d\theta d\phi = 16\pi \int_0^\pi \sin^2(\phi) d\phi = 8\pi^2$$ 
A: The mass is over a spherical surface. So, it does not appear right to integrate along the radial direction, as done in your setup.
Note that the density is symmetrical around the $z$-axis. Therefore, the mass integration can be performed in the two-dimensional coordinates $(\rho, z)$, where $\rho=\sqrt{x^2+y^2}$ and $r^2=\rho^2+z^2=4$.
The integral can be set up in its corresponding polar form, with $\rho=r\cos\theta$,
$$m=\int_{-\pi/2}^{\pi/2} \sigma(\rho) \cdot 2\pi\rho \>rd\theta =2\pi r^3 \int_{-\pi/2}^{\pi/2} \cos^2\theta d\theta=\pi^2r^3=8\pi^2$$
