Question

Problem 4 Evaluate the line integrals (a) (10 points) y da 2ax dy, where C is the curve r(t) (2t + 1) i+ 3t2 j, 0t 1. (b) (10 points) (ryz) ds, where C is the line segment from the point (2, 1,0) to the point (4,3,6) (c) (10 points) F.dr,where F is the vector field F(x, y) = yi - rj and C is the curve given by r(t) t2i+ (1+ t3) j, 0 < t < 1.

please respond with explanations for each step. thank you

Answer 1

I have explained methods, hope you will love them; there might be calculating errors if there are, please correct them.

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Sol a) ydr 2xdy L18 1(1+1 = x : axauM x =2ttl, y 3t da 2dtdy6tdt ydadya ettiiat 2dt+2 (2tt)6tdt C 4 4tttt 12t(2tt0|dt 24t+24tt 6t+ 124tt2t)d£ 4 =24t+24tt30tt 12t) d 24 24 4 12 2- 144

Whexe c ine txo (2,1,0) to (4,3,6) (y,2,) (x22,22) The eqtaf Tine in 3D 4-9, - Z- y2-y L- Z2-Z 2-2 -y-T 3- - Z-0>t 4-2 32t+2,42tt zt3 osts dx= 2dt, dy 2dt, dz= 6dt dz 6 e dx 2 dt 2 dt dt ds Tdz 12 dx dt /dy dt dt 4P dr 2 dt di211 dt ) txy+ zids = f [et +2{2+) +ét] 2dt 412t12)dt C 12 42 4 2 2 f 211 2 3 56 11

Fege No Deie: F.dx whexe F(x) =y -ズ」 ts ost it1tt3j. C 2. ty=1tt d(2+1+3tdt F(1tt1-t F dx-(2t (tt stdt - 2t fi+21+ 2. F.dx(2 tt4t+2t-3t) dt 4 PL Fdx ttat2t-stdt 2 3 5 4