Answer to Question #253106 in Calculus for Anuj

Question #253106
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.
1
Expert's answer
2021-10-26T14:15:52-0400

"z=3-x^2-y^2,x\\ge2"


a)

curve C: "x^2+y^2=1"




b)

"\\iint_S=\\iint_S curl\\vec{F}\\cdot d\\vec{S}=\\iint_S curl\\vec{F}\\cdot\\vec{n}dS"


"curl\\vec{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 curl\\vec{F}\\cdot d\\vec{S}=\\iint_D(1,0,-2)(-2x,-2y,-1)dA=\\iint_D(2-2x)dA"

"x=rcos\\theta,y=rsin\\theta"

"0\\le \\theta\\le 2\\pi,0\\le r\\le 1"


"\\iint_S curl\\vec{F}\\cdot d\\vec{S}=\\intop^{2\\pi}_0\\int^1_0(2-2rcos\\theta)r drd\\theta="


"=\\intop^{2\\pi}_0(1-2cos\\theta\/3) d\\theta=(\\theta-\\frac{2}{3}sin\\theta)|^{2\\pi}_0=2\\pi"


Stokes’ Theorem:

"\\int_C \\vec{F}\\cdot d\\vec{r}=\\iint_S curl\\vec{F}\\cdot d\\vec{S}"


"\\vec{r}(t)=(cost,sint,2),0\\le t\\le 2\\pi"


"\\int_C \\vec{F}d\\vec{r}=\\int^{2\\pi}_0\\vec{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 curl\\vec{F}\\cdot d\\vec{S}==\\int^{2\\pi}_0(-2sin^2t+6cost)dt=(-\\frac{sin2x-2x}{2}+6sint)|^{2\\pi}_0=2\\pi"


XY-projection of the surface:

"x^2+y^2=1"





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