Answer to Question #246038 in Calculus for Njabulo

Question

11. (a) State Stokes’ Theorem for converting a flux integral over a bounded surface to a line integral over the
curve that bounds the surface. (b) Consider the surface
S =

(x, y, z) | z = x
2 + y
2
; z ≤ 9

.
Sketch the surface S in R
3 and show the curve that bounds S on your sketch. Then use Stokes’ Theorem
to evaluate the flux integral
Z Z
S
(curl F)

Answer 1

a)

$\iint_S curlF\cdot nds=\oint_C F\cdot dr$

where F is C¹- -vector field defined on an open region in R³ containing S, and S and C are piecewise smooth, and n is the outward normal of the surface and C is positively oriented (anti-clockwise).

b)

$C=\{(x,y,z):z=9,x^2+y^2=9\}$

 parametrise C:

$r(t)=(3cost,3sint,9)$ , $0\le t\le 2\pi$

$r'(t)=(-3sint,3cost,0)$

$F(r(t))=F(3cost,3sint,9)=(3sint,-3cost,9)$

Using Stokes’s Theorem:

$\iint_S curlF\cdot nds=\oint_C F\cdot dr=\int^{2\pi}_0F(r(t))\cdot r'(t)dt=$

$=\int^{2\pi}_0(3sint,-3cost,9)\cdot(-3sint,3cost,0)dt=$

$=\int^{2\pi}_0(-9sin^2t-9cos^2t)dt=-18\pi$