Question #267487

Evaluate the line integral

āˆ«š’–(š‘„, š‘¦, š‘§) Ɨ ā…†š’“ ,

where š’–(š‘„, š‘¦, š‘§) = (š‘¦^2 , š‘„, š‘§) and the curve š‘Ŗ is described by š’› = š‘¦ = š‘’ š‘„ with š‘„ ∈ [0,1]. 



Expert's answer

parameter for of C,

x=t, y=et ,z=et .

r(t)=t i+ et j+et k

=> dr=(i + et j + et k)dt

Then,

∫01u.dr=∫01(e2t,t,et).(1,et,et)dt=∫01(2e2t+tet)dt=[e2t+tetāˆ’et]01=e2+eāˆ’eāˆ’1āˆ’0+1=e2\int_0^1 u.dr\\ =\int_0^1 (e^{2t},t,e^{t}).(1,e^t,e^t)dt\\ =\int_0^1 (2e^{2t}+te^{t})dt\\ =[ e^{2t}+te^{t}-e^t]_0^1\\ =e^2+e-e-1-0+1\\ =e^2


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