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**HallsofIvy** You are **told** the figure is a cone. In the x,z plane (y= 0), the equation reduces to $\displaystyle x^2= z^2$ 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 $\displaystyle \theta= 0$ 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 $\displaystyle x= cos(\theta)$, $\displaystyle y= sin(\theta)$, $\displaystyle z= 1$ so that $\displaystyle dx= -sin(\theta)d\theta$, $\displaystyle dy= cos(\theta)d\theta$, and $\displaystyle dz= 0$

On the surface, $\displaystyle z= \sqrt{x^2+ y^2}$, so the position vector of a point on the cone is $\displaystyle \vec{X}= x\vec{i}+ y\vec{j}+ z\vec{k}= x\vec{i}+ y\vec{j}+ \sqrt{x^2+ y^2}\vec{k}$ which, in cylindrical coordinates, is $\displaystyle r cos(\theta)\vec{i}+ r sin(\theta)\vec{j}+ r\vec{k}$.

The derivatives with respect to r and $\displaystyle \theta$ are $\displaystyle \vec{X}_r= cos(\theta)\vec{i}+ sin(\theta)\vec{j}+ \vec{k}$ and $\displaystyle \vec{X}_\theta= -rsin()\theta)\vec{i}+ rcos(\theta)\vec{j}$ 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 $\displaystyle -r cos(\theta)\vec{i}- r sin(\theta)\vec{j}+ r\vec{k}$ and so the "vector differential of surface area is $\displaystyle (-r cos(\theta)\vec{i}- r sin(\theta)\vec{j}+ r\vec{k})drd\theta$.

Calculate $\displaystyle \nabla\times \vec{f}$, put it in polar coordinates with $\displaystyle x= r cos(\theta)$, $\displaystyle y= r sin(\theta)$, and $\displaystyle z= r$, take the dot product with $\displaystyle d\vec{S}$ and integrate. r will run from 0 to 1, $\displaystyle \theta$ from 0 to $\displaystyle 2\pi$.

The actual integrals are almost trivial.