You mean "that part of the plane 2x+3y+6z=12 which is located in the first octant". The additional "of the plane 2x+ 3y+ 6z= 12" confused me!

When x and y are 0, that becomes 6z= 12 or z= 2 so one corner of the triangle is (0, 0, 2). When y and z are 0, that becomes 2x= 12 or x= 6 so (6, 0, 0) is another corner of the triangle.

And when x and z are 0, that becomes 3y= 12 or y= 4 so (0, 4, 0) is the third corner. Yes, the unit normal to the plane is(2i+ 3j+ 6k)/7(note the parentheses1)

The "sqrt" in "sqrt(...)dx"isthat "7", the length of the vector.

Another way of looking at surface integrals, which I prefer, is this: We can write any curve, in an xyz coordinate system, as x= f(s,t), y= g(s,t), z= h(s, t) where s and t are the parameters.

And we can write that as the vector equation . The two derivatives, and are vectors lying in the tangent plane to the surface so that is normal to the surface. Further, we have .

Here, our surface is the plane 2x+ 3y+ 6z= 12 which we can write as z= 2- (1/3)x- (1/2)y and write x= 3s, y= 2t, z= 2- s- t. The "position vector" or a point on the plane would be . The derivatives are and . The cross product of those two vectors (positive z component) is so that .

The integrand vector function is so

Projecting 2x+3y+ 6z= 12 into the xy-plane (z= 0) gives 2x+ 3y= 12 or y= 4- (2/3)x while x runs from 0 to 6. Since s= x and t= y, The integral will be