Consider an empty thin ice cream cone standing with its tip at the origin. The height of the cone is 3 and the radius of the top is 6. Find the center of gravity of the cone by following the steps below. Assume the density of the cone is constant 1.
a. We parametrize the cone $S$ with $s(r,t)=(r\cos(t),r\sin(t),$ $)$ where $0\le t \le 2\pi$ and $0\le r \le 6$.

b. The partial derivatives are $s_r(r,t) =$ (,,) and $s_t(r,t) =$ (,,).

c. The cross product is $s_r(r,t)\times s_t(r,t) =$ (,,).

d. The norm of the cross product is $\| s_r(r,t)\times s_t(r,t) \|=$ .

e. $m = \int\int_S 1 dS =$

f. $m_z = \int\int_S z dS =$

g. The center of gravity is $(0,0,\bar z)$ where $\bar z=\frac{m_z}{m} =$ .