Question

Ja Problems 6 through 10, use Stokes' theorem to evaluate F.Tds. OF=3yi - 2xj + 3yk; C is the circle x2 + y2 = 9, Z = 4. oriented counterclockwise as viewed from above. 1.F=2zi+xj+3yk; C is the ellipse in which the plane z = x meets the cylinder x? + y2 = 4, oriented counterclockwise as viewed from above. & F= yi+zj+xk; C is the boundary of the triangle with ver- tices (0,0,0), (2,0,0), and (0, 2, 2), oriented counterclock- wise as viewed from above. 9. F= (y- x,x-z, x-y); C is the boundary of the part of the plane x + 2y + z = 2 that lies in the first octant, oriented counterclockwise as viewed from above. 10. F=y41+z?j + x?k; C is the intersection of the plane z = y and the cylinder x2 + y2 = 2y, oriented counterclockwise as viewed from above.

#10

Answer 1

10. F = 42; +229 + x2 z 2-4 and x² + y² = 24 Stokes theorem & Firds - SS exter as fon dady Inokl Non 2 = y Now ñ 2 -4 = 0 as = D(z-y) = oi g + ² IS & oi-7 tk IISI S0²E11 2+(112 oingt 10.kl oi-gt 12 Now DX on 2 Sy 09 z2 ? OZ x2 = (-22) - (2x - 0) + (0-24) Ř - - 221-2x 7 – 24 F

sxes – (-22,- 2x - 2y) - (0 / 1 / 22 √2 6F.Tds = 552625%) dxdy - 2 ss (x-4) ax ay Now x² + y2 = 24 & (x-6) 2+ (4-1)2 = using polare coordinate set uz & cos o y = r sina & cos²o +r2 sino = 2rsino की । r2 = arsind se o, a sino linult of our & a sino and ouoct

I asino (rcoso - rsino) sardo & F.Tds - 2 eru r 2 sino -21 Caso-sino) 8²ardo 00 asino 21 (coso 3 O 8 (coso-sino) sino do 1+ COSO - co wloo No 8 (coso sino-sino) do 1 -Cros COIM VORO (ccoso sin 30- (1-10820²) do ( 1+ cos220 – 2 cos20)) do 4 Els Secoso sino - 5 (1 + i + (0340 – 2.60520)) do Tag ( Signage -- (0+ 9+ stano - Sinzo) ST -Io + 0 + sin 40 - (sinn á (r+ p + sincast) - siner - (O)) =16C-(39) 1-20 | Answer O F. Tas ritas - Answer 2