Question

Compute the vector line integral  \( \int F.dr \) , where \( F \) and C are as indicated.

(a). \( F(x,y)=(siny,x) \) ; C : \( r(t)=(t^2-1,t),t\in [o,\pi] \)

(b). \( F(x,y)=(x,y+1) \) ; C : \( r(t)=(1-sint,1-cost),t\in [0,2\pi] \)

Answer 1

solution

(a). \( F(x,y)=(siny,x) \) ; C : \( r(t)=(t^2-1,t),t\in [o,\pi] \)

we have \( r(t)=(t^2-1,t)=>r'(t)=(2t,1) \) and \( F(x,y)=(siny,x)=>F(r(t))=(sint,t^2-1) \)

\( =>\int_CF.dr=\int_a^bF(r(t)).r'(t)dt=\int_0^\pi(sint,t^2-1).(2t,1)dt \)                       \( =\int_0^\pi(2tsint+t^2-1)dt=\frac{\pi^3+3\pi}{3} \)

Threfore, \( \int_CF.dr=\int_a^bF(r(t)).r'(t)dt=\frac{\pi^3+3\pi}{3}=13.47 \)

(b). \( F(x,y)=(x,y+1) \) ; C : \( r(t)=(1-sint,1-cost),t\in [0,2\pi] \)

we have \( r(t)=(1-sint,1-cost)=>r'(t)=(-cost,sint) \) and \( F(x,y)=(x,y+1)=>F(r(t))=(1-sint,2-cost) \)\( =>\int_CF.dr=\int_a^bF(r(t)).r'(t)dt=\int_0^{2\pi}(1-sint).(-cost,sint)dt \)                       \( =\int_0^{2\pi}(-cost+2sint)dt=0 \)

Therfore, \( \int_CF.dr=\int_a^bF(r(t)).r'(t)dt=0 \)

answer

(a).Thefore, \( \int_CF.dr=\int_a^bF(r(t)).r'(t)dt=\frac{\pi^3+3\pi}{3}=13.47 \)

(b), Threfore, \( \int_CF.dr=\int_a^bF(r(t)).r'(t)dt=0 \)