Here's a general concept: if you write the surface as

, then the two derivatives,

and

are tangent to the surface and their cross product (called the "fundamental vector product" of the surface),

, is perpendicular to the surface and its length gives the "differential of surface area":

.

Yes, you can use

,

, x= x with

and x as parameters u and v. That is,

so

and

so that the "fundamental vector product" is

.

Its length is, of course, 1 so

. The function to be integrated is

. In the first octant, both y and z are positive so

goes from 0 to

and, of course, x goes from 0 to 1. The integral is

.

By the way, if you click on a mathematical expression in any post, you can see the LaTex code used.