Answer to Question #238272 in Calculus for Anuj

Question

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. 

Answer 1

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

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:

$x^2+y^2=1$