(Q) Find the area of the surface cut from the paraboloid x^2+y+z^2 = 2 by the plane y=0.
The Attempt at a Solution
The unit normal vector in this case will be j. Moreover, the gradient vector will be
sqrt(4x^2+4z^2+1). And the denominator which is the dot product of the gradient vector and j is 1 so we need not bother about that.
So the double integral will be that of sqrt (4x^2+4z^2+1) but since y=0, it means that x^2+z^2 = 2 so sqrt (4x^2+4z^2+1) becomes 3.
The problem is that the solution manual does not do this. It retain the integral as sqrt (4x^2+4z^2+1) and uses polar co-ordinates which I can do but I don't understand why we cannot substitute. Please explain.
Thank-you very much for the time and effort!!!