On the "curved part" of the cone, use cylindrical coordinates so that

,

, and

. That is, use "position vector"

. Then

and

. The cross product of those vectors is

where I have chosen the order of the product to give the "outward pointing normal". Its length,

gives the surface integral:

For the flat surface, that is just

,

, a disk of radius h, and its area is just

.

By the way, you don't really need to integrate to find the curved surface area of a cone. Given a cone with height h and base radius R, cut it along a straight line from base to vertex. Then "flatten" it. (You can do this- any surface with at least one straight line in the surface through every point- a "developable surface"- can be flattened). That forms part of a disk of radius

, the "slant height", and area

. The circumference of that larger disk would be

of course, while the "circumference" of our cone was only

. That is a ratio of

. That means that the curved surface area of our cone must be

. Here, R= h so that is

.