- #1

- 92

- 0

## Homework Statement

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

## Homework Equations

## 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!!!