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1.) Evaluate the line integral over y=x^2 (0 F=(2x+2y)i + (x^2 -y)j
my actual answer came out to be 5/3... but why does our instructor say that the answer should come out to be 1. Did he do something wrong???

b.) Evaluate the line integral F =yi + x^2j over y=(1/2)x^2 (0 Once again I ended up with -6 when performed myself.

2007-09-29 08:28:10 · 1 answers · asked by jack86 1 in Science & Mathematics Mathematics

1 answers

x=t and y=t^2
F = (2t +2t^2) i +(t^2-t^2) j so F= (2t+2t^2)i
Int(0,1) (2t+2t^2)dt = 2(t^2/2+t/3/3) (0.1) = 2(1/2+1/3) =5/3
you are right
2) x=t
y=1/2 t^2 dy = t dt
F=1/2t^2 i +t^2 j
Int Pdx +Q dy (P(x,y) = y and Q(x,y) = x^2)
= Int(0,2) 1/2 t^2 dt +t^3 dt = t^3/6+t^4/4= 4/3+4=16/3

2007-09-29 08:57:40 · answer #1 · answered by santmann2002 7 · 0 0

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