Let vec$(F)=(3z+3x^3)vec(i)+(6y+6z+6sin(y^3))vec(j)+(3x+6y+6e^{z^3})vec(k).$

$(a)$ Find curlvec$(F).$

curlvec$(F)=$

$(b)$ What does your answer $to$ part $(a)$ tell you about ${\displaystyle \underset{C}{\overset{\phantom{}}{}}}$vec$(F)*dvec(r)$ where $Cis$ the circle

$(x-30{)}^2+(y-30{)}^2=1in$ the $xy-$plane, oriented clockwise?

${\displaystyle \underset{C}{\overset{\phantom{}}{}}}$vec$(F)*dvec(r)=$

$(c)IfCis$ any closed curve, what can you say about ${\displaystyle \underset{C}{\overset{\phantom{}}{}}}$vec$(F)*dvec(r)?$

${\displaystyle \underset{C}{\overset{\phantom{}}{}}}$vec$(F)*dvec(r)=$

$(d)$ Now let $Cbe$ the half circle $(x-30{)}^2+(y-30{)}^2=1in$ the $xy-$plane with $y>30,$

traversed from $(31,30)to(29,30).$ Find ${\displaystyle \underset{C}{\overset{\phantom{}}{}}}$vec$(F)*dvec(r)by$ using your result from $(c)$ and

considering $C$ plus the line segment connecting the endpoints $ofC.$

${\displaystyle \underset{C}{\overset{\phantom{}}{}}}$vec$(F)*dvec(r)=$