(Q) Find the area of the portion of the surface x^2 - 2z = 0 that lies above the triangle bounded by the lines x = sqrt(3), y = 0, and y = x in the xy-plane.
The Attempt at a Solution
The know how to find the gradient vector. The part where I am facing a problem is in finding the unit normal vector. the solution manual simply states that it is equal to k.
However, I thought that since the surface is defined by an equation that contains only x and z, the normal vector must be j.
I've been facing lots of problems in finding the normal vector lately. Can someone please explain to me the procedure to do so?
Thank-you very much!!!