# Answer in Calculus for Anuj #238272

Consider the surface S = (x, y, z) ∈ R 3 | z = 3 − x 2 − y 2 ; z ≥ 2 . Assume that S is oriented upward and let C be the oriented boundary of S. (a) Sketch the surface S in R 3 . Also show the oriented curve C and the XY-projection of the surface S on your sketch. (2) (b) Let F (x, y, z) = (2y, 3z, 4y). Evaluate the flux integral Z Z S (curl F) · n dS by i. determining curl F and the upward unit normal n of S and using the formula (17.2) on p. 104 of Guide 3 (5) ii. Using Stokes’ Theorem, convert the given flux integral to a line integral.

“z=3-x^2-y^2,xge2”

a)

curve C: “x^2+y^2=1”

b)

“iint_S=iint_S curlvec{F}cdot dvec{S}=iint_S curlvec{F}cdotvec{n}dS”

“curlvec{F}=(R_y-Q_z)i+(P_z-R_x)j+(Q_x-P_y)k=i-2k”

“vec{n}=frac{-nabla f}{||-nabla f||}”

“f(x,y,x)=z+x^2+y^2-3”

“nabla f=(2x,2y,1)”

“iint_S curlvec{F}cdot dvec{S}=iint_D(1,0,-2)(-2x,-2y,-1)dA=iint_D(2-2x)dA”

“x=rcostheta,y=rsintheta”

“0le thetale 2pi,0le rle 1”

“iint_S curlvec{F}cdot dvec{S}=intop^{2pi}_0int^1_0(2-2rcostheta)r drdtheta=”

“=intop^{2pi}_0(1-2costheta/3) dtheta=(theta-frac{2}{3}sintheta)|^{2pi}_0=2pi”

Stokes’ Theorem:

“int_C vec{F}cdot dvec{r}=iint_S curlvec{F}cdot dvec{S}”

“vec{r}(t)=(cost,sint,2),0le tle 2pi”

“int_C vec{F}dvec{r}=int^{2pi}_0vec{F}(vec r(t))cdot vec{r}'(t)dt”

“vec{F}(vec r(t))=(2sint,6,4sint)”

“vec{r}'(t)=(-sint,cost,0)”

“iint_S curlvec{F}cdot dvec{S}==int^{2pi}_0(-2sin^2t+6cost)dt=(-frac{sin2x-2x}{2}+6sint)|^{2pi}_0=2pi”

XY-projection of the surface:

“x^2+y^2=1”

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