2. Convert to cartesian coordinates and identify the surface. (a) (5 pts) $2 r^{2}+\rho^{2}=4$ (b) (5 pts) $\rho=3 \sec \phi$

Step 1 ：Given the equation in spherical coordinates $2 r^{2}+ ho^{2}=4$, we can convert it to cartesian coordinates using the formulas $x = \rho \sin\phi \cos\theta$, $y = \rho \sin\phi \sin\theta$, $z = \rho \cos\phi$, $r = \sqrt{x^2 + y^2}$ and $\rho = \sqrt{x^2 + y^2 + z^2}$.

Step 2 ：Substituting these into the given equation, we get $3x^2 + 3y^2 + z^2 = 4$.

Step 3 ：This is the equation of an ellipsoid centered at the origin with semi-axes of lengths $\sqrt{\frac{4}{3}}$ along the $x$ and $y$ axes and $2$ along the $z$ axis.

Step 4 ：\(\boxed{\text{The surface is an ellipsoid centered at the origin with semi-axes of lengths } \sqrt{\frac{4}{3}} \text{ along the } x \text{ and } y \text{ axes and } 2 \text{ along the } z \text{ axis.}}\)

Step 5 ：Given the equation in spherical coordinates $\rho=3 \sec \phi$, we can convert it to cartesian coordinates using the formulas $x = \rho \sin\phi \cos\theta$, $y = \rho \sin\phi \sin\theta$, $z = \rho \cos\phi$, $\rho = \sqrt{x^2 + y^2 + z^2}$ and $\sec \phi = \sqrt{x^2 + y^2 + z^2}/z$.

Step 6 ：Substituting these into the given equation, we get $z = 3, z = -\sqrt{x^2 + y^2}, z = \sqrt{x^2 + y^2}$.

Step 7 ：This represents a cone with vertex at the origin and axis along the $z$-axis, and a plane $z = 3$.

Step 8 ：\(\boxed{\text{The surface is a cone with vertex at the origin and axis along the } z \text{-axis, and a plane } z = 3.}\)