Question #183538

Evaluate LaTeX: \int_cF.dr\:\: where LaTeX: F\left(x,y,z\right)=xzi-yzkF(x,y,z)=xzi−yzk and c is the line segment from (3,0,1) to (-1,2,0)


1
Expert's answer
2021-05-07T12:30:50-0400

Given,


F(x,y,z)=xziyzk Nd line segment c from (3,0,1)(1,2,0)F(x,y,z)=xz \textbf i−yz \textbf k \ Nd \ line \ segment \ c\ from\ (3,0,1)\to(-1,2,0)

Let,


r(t)=(1t)<3,0,1>+t<1,2,0>=<34t,2t,1t>r(t) =(1-t)<3,0,1>+t<-1,2,0>=<3-4t,2t,1-t>

Differentiate with respect to t,


dr=(4i+2jk)dtdr=(-4\textbf i+2\textbf j-\textbf k)dt

Now,


F(r(t))=(34t)(1t)i2t(1t)kF(r(t))=(3-4t)(1-t)i-2t(1-t)k

F(r(t))=(3t27t+3)i+(2t22t)kF(r(t))=(3t^2-7t+3)\textbf i+(2t^2-2t)\textbf k

Therefore,


cF.dr=[(3t27t+3)i+(2t22t)k][4i+2jk]dt\int_cF.dr=[(3t^2-7t+3)\textbf i+(2t^2-2t)\textbf k]\cdot[-4\textbf i+2\textbf j-\textbf k]dt\\

=014(3t27t+3)(2t22t)dt=0114t2+30t12dt=[143t3+15t212t]01=53=\int_{0}^{1}-4(3t^2-7t+3)-(2t^2-2t)dt\\ =\int_{0}^{1}-14t^2+30t-12dt\\ =[-\frac{14}{3}t^3+15t^2-12t]_{0}^{1}\\ =-\frac{5}{3}

Thus, the required answer is 53\frac{5}{3} .


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