Use vec$(F)=x^3$vec$(i)+y^3$vec$(j)+z^3$vec$(k)$ and the Divergence Theorem $to$ find the flux $of$ vec$(F)$ out $of$ each $of$ the closed surfaces below. Set $up$ the

triple integral needed.

$(a)Sis$ the closed cone $z=\sqrt[\phantom{2}]{x^2+y^2},$ with $0z4,$ oriented outward.

$\{\begin{array}{ccc}\text{:}{\displaystyle \underset{S}{\overset{\phantom{}}{}}}\text{v}\text{e}\text{c}& (F)*dvec(A){\displaystyle \underset{\text{l}\text{a}\text{r}\text{r}}{\overset{\phantom{}}{}}}{\displaystyle \underset{V}{\overset{\phantom{}}{}}}(\frac{?}{\text{b}\text{a}\text{r}}& (V))dzdrd\end{array}$

$(b)Sis$ the "snow cone" bounded $byan$ upper hemisphere $of$ radius $5$ and a standard cone, oriented outward.

$\{\begin{array}{cc}\text{:}{\displaystyle \underset{S}{\overset{\phantom{}}{}}}\text{v}\text{e}\text{c}& (F)*dvec(A){\displaystyle \underset{}{\overset{}{}}}\sqrt[\phantom{2}]{}({|}_{}^{})ddd\end{array}$

$(c)Sis$ the closed cylinder $of$ radius $8$ and height $4,$ centered $on$ the positive $z-$axis $so$ that the bottom $of$ the cylinder $ison$ the plane $z$

$={\displaystyle \underset{\text{d}\text{a}\text{r}\text{r}}{\overset{\phantom{}}{}}}{\displaystyle \underset{AA}{\overset{\phantom{}}{}}}{\displaystyle \underset{\text{d}\text{a}\text{r}\text{r}}{\overset{\phantom{}}{}}}()dzdrd$