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)


Expert's answer

Given,


F(x,y,z)=xzi−yzk 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)=(1−t)<3,0,1>+t<−1,2,0>=<3−4t,2t,1−t>r(t) =(1-t)<3,0,1>+t<-1,2,0>=<3-4t,2t,1-t>

Differentiate with respect to t,


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

Now,


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

F(r(t))=(3t2−7t+3)i+(2t2−2t)kF(r(t))=(3t^2-7t+3)\textbf i+(2t^2-2t)\textbf k

Therefore,


∫cF.dr=[(3t2−7t+3)i+(2t2−2t)k]⋅[−4i+2j−k]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\\

=∫01−4(3t2−7t+3)−(2t2−2t)dt=∫01−14t2+30t−12dt=[−143t3+15t2−12t]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} .


LATEST TUTORIALS
APPROVED BY CLIENTS