You are

**told** the figure is a cone. In the x,z plane (y= 0), the equation reduces to

or the two lines z= x and z=-x. In the y,z plane (x= 0), it is z= y and z= -y. Use those to sketch it. Alternatively, use the cylindrical coordinates, z= r. Draw the line z= r in the

plane and rotate it around the z-axis.

??? There is no

**path** mentioned in this problem! Do you mean the boundary at z= 1? That is a circle with center at (0, 0, 1), in the z= 1 plane, with radius 1. It is given by

,

,

so that

,

, and

On the surface,

, so the position vector of a point on the cone is

which, in cylindrical coordinates, is

.

The derivatives with respect to r and

are

and

and are vectors in the tangent plane at each point.

Their cross product, the "fundamental vector product" for the surface, oriented in the positive z direction, is

and so the "vector differential of surface area is

.

Calculate

, put it in polar coordinates with

,

, and

, take the dot product with

and integrate. r will run from 0 to 1,

from 0 to

.

The actual integrals are almost trivial.