Surface Area and Surface Integrals: x^2+y+z^2 = 2

[CalculusLover]

(Q) Find the area of the surface cut from the paraboloid x^2+y+z^2 = 2 by
the plane y=0.

My attempt:

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!!!

 

[tkhunny]

y = 0 defines your limits, not your integrand. You've turned your paraboloid into a cyllinder!

x^2 + z^2 = 2 ONLY in the x-z plane (y = 0).
At y = 2, x^2 + z^2 = 0
And, of course, everything in between.