Answer to Question #183546 in Calculus for Maureen wariara

Question

Evaluate   where   

F(x,y,z)=xzi−yzk

  and c is the line segment  from (3,0,1) to (-1,2,0)

Answer 1

The line segment from $(3,0,1) to (-1,2,0)$ is,

$r(t)=\langle 3,0,1\rangle+t\langle -4,2,-1\rangle$

$=\langle 3,0,1\rangle+\langle -4t,2t,-t\rangle$

$=\langle 3-4t,2t,1-t\rangle$

where, $0<t<1$

The line integral is evaluated as,

$\int_{C}F(x,y,z)ds$

$=\int_{C}F(r(t))\cdot r'(t)dt$

$=\int_{0}^{1}\langle (3-4t)(1-t),0,-2t(1-t) \rangle\cdot \langle -4,2,-1 \rangle dt$

$=\int_{0}^{1}(-4(3-7t+4t^2)+2t(1-t)) dt$

$=\int_{0}^{1}(-12+28t-16t^2+2t-2t^2) dt$

$=\int_{0}^{1}(-12+30t-18t^2) dt$

$=\lbrack -12(t)+30(\frac{t^2}{2})-18(\frac{t^3}{3})) \rbrack_{0}^{1}$

$=-12+15-6$

$=-3$

Therefore, the line integral is $\int_{C}F(x,y,z)ds=-3$