Question #246038
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)
1
Expert's answer
2021-10-05T11:26:48-0400

a)

ScurlFnds=CFdr\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,x2+y2=9}C=\{(x,y,z):z=9,x^2+y^2=9\}

 parametrise C:

r(t)=(3cost,3sint,9)r(t)=(3cost,3sint,9) , 0t2π0\le t\le 2\pi

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

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


Using Stokes’s Theorem:


ScurlFnds=CFdr=02πF(r(t))r(t)dt=\iint_S curlF\cdot nds=\oint_C F\cdot dr=\int^{2\pi}_0F(r(t))\cdot r'(t)dt=


=02π(3sint,3cost,9)(3sint,3cost,0)dt==\int^{2\pi}_0(3sint,-3cost,9)\cdot(-3sint,3cost,0)dt=


=02π(9sin2t9cos2t)dt=18π=\int^{2\pi}_0(-9sin^2t-9cos^2t)dt=-18\pi


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