Consider an empty thin ice cream cone standing with its tip at the origin. The height of the cone is $7$ and the radius of the top is $5.$ 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$)=($rcos$($t$),$rsin$($t$),)$ where $0$t$2$ and $0$r$5.$

b$.$ The partial derivatives are sr$($r$,$t$)=(,,)$ and st$($r$,$t$)=(,,).$

c$.$ The cross product is sr$($r$,$t$)$st$($r$,$t$)=(,,).$

d$.$ The norm of the cross product is $$sr$($r$,$t$)$st$($r$,$t$)=.$

e$.$ m$=$S$1$dS$=$

f$.$ mz$=$SzdS$=$

g$.$ The center of gravity is $(0,0,$z$)$ where z$=$mzm$=.$