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### 1 Answer

a. Let S be the given surface with its boundary C, which path. Then Stoke's Theorem relates a line integral and surface integral of some vector field F i.e.,

`int_C vecF.d vecr=int int_S curlvecF.dvec S`

We have given vector field F=<yz,-xz,z^3> and surface

S={(x,y,z): z=-sqrt(x^2+y^2) and `-2<=z<=0}`

We have

`-2=-sqrt(x^2+y^2)`

`x^2+y^2=4`

Let transform this equation in parametric form

`x=rcos(t)`

`y=rsin(t)`

`r^2cos^(t)+r^2sin^2(t)=4`

`=>r^2=4`

`=> r=2`

Thus

`vecr(t)=2cos(t)vec i+2sin(t)vecj-2veck`

`dvecr=(-2sin(t) vec i +2cos(t) vecj) dt`

`vecF=(2sin(t)(-2)) veci -(2cos(t))(-2)vec j+ (-2)^3 vec k`

`=-4sin(t)veci+4cos(t) vecj -8vec k`

`vecF.dvecr=(8sin^2(t)+8cos^2(t))dt=8dt`

`int_CvecF.dvecr=int_0^pi8dt=8pi`

`curlF=(z^4+xyz-x^2z+3xz^3)veci+(z^4+y^2z-xyz+3yz^3)vecj+(xz^2-yz^2)veck`

```dvecS=(gradg)/||gradg|| ||gradg|| dA`

`g(x,y,z)=z+sqrt(x^2+y^2)`

`gradg=x/sqrt(x^2+y^2) veci+y/sqrt(x^2+y^2) vecj+vec k`

`||gradg||=sqrt((x/sqrt(x^2+y^2))^2+(y/sqrt(x^2+y^2))^2+1)`

`=sqrt(2)`

`dA=dx.dy`

Thus substitute these values in surface integral and evalute. We will

get 8pi.