line integral problem please?!!?

Question

Evaluate the line integral if F(x,y,z)=e^x i +e^y j + e^z k


-R(t)= t i t^2 j + t^3 k

0

-Please do WRITE ALL THE SOLUTIONS, for me to understand well.

-extended thanks!!! :-)

Answer 1

The value of the line integral will be the integral of F.dR along the curve C, from t = 0 to t = 2.

Now dR = dt i + 2tdt j + 3t²dt k

and F along C on R(t) = e^t i + e^(t²) j+ e^(t³) k.

Therefore: F.dR = e^tdt + e^(t²)*2tdt + e^(t³)*3t²dt

= [e^t + e^(t²)*2t + e^(t³)*3t²]dt

So that the value of the line integral = ∫[e^t + e^(t²)*2t + e^(t³)

*3t²]dt from t = 0 to t = 2,

= ∫[e^t + e^(t²)*2t + e^(t³)*3t²]dt = e^t + e^(t²) + e^(t³) + K (constant if integration)

And its value from t = 0 to t = 2 will be

= e² + e^4 + e^8 - 1 - 1 - 1 = e² + e^4 + e^8 - 3 ≈ 3039.95

Answer 2

so this mapping will be a vector function to a vector function, you'll pry want to use this formula

integral F around a curve C is given by

integral from 0 to 2 of F(R(t)) dot R'(t) dt

parameterizing shouldn't be particularly bad
F(R(t)) = ( e^t, e^t^2, e^t^3)

R'(t) = (1, 2t, 3t^2)

the dot product is
e^t + 2te^t^2 + 3t^2 e^t^3

integrals are quite easy, some nice u subs

int e^t = e^t,
int 2te^t^2 = e^t^2
int 3t^2 e^t^3 = e^t^3

now evaluate from t=0 to 2

e^2+e^4+e^8 - 3

Answer 3

x=t,y=t^2,z=t^3 as R(t)= ti+t^2j+t^3k and ds=dxi+dyj+dzk
Int F*ds where ds= dt*i+2tdt*j+3t^2dt*k
so the integral becomes
Int(0,2) e^tdt +2te^(t^2) dt +3t^2(e^t3)dt= (e^2-1)+(e^4-1)+(e^8-1)