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.