Let C consist of straight lines joining #(3,1,2), (2,1,5),# and #(1,0,1)#. evaluate the integral of #phi(x,y,z)=xyi+yzj+xzk# around C?

1 Answer
May 7, 2018

#55/2#

Explanation:

#bb phi(x,y,z)=xy \ bb hat i+yz \ bb hat j+xz \ bb hat k#

  • <3,1,2> to <2,1,5>

The path is:

#bb r_1 = langle 3,1,2 rangle + lambda langle-1,0,3 rangle, qquad d bb r_1 = langle-1,0,3 rangle d lambda #

The integral is:

#I_1 = int_(bb r_1) \ ((xy),(yz),(xz))\ cdot \ d bb r_1#

#= int_0^1 \ (((3 - lambda)(1)),((1)(2 + 3 lambda)),((3 - lambda)(2+3 lambda)))\ cdot ((-1),(0),(3) ) \ d lambda #

#= int_0^1 \ (lambda - 3) + 0 + 3(3 - lambda)(2+3 lambda) \ d lambda #

#= int_0^1 \ -9 lambda^2 + 22 lambda + 15 \ d \ lambda = 23#

  • <2,1,5> to <1,0,1>

#bb r_2 = langle 2,1,5 rangle + lambda langle-1,-1,-4 rangle, qquad d bb r_2 = langle-1,-1,-4 rangle d lambda #

# I_2 = int_0^1 \ (((2-lambda)(1- lambda)),((1- lambda)(5 - 4 lambda)),((5 - 4 lambda)(2 - lambda)))\ cdot ((-1),(-1),(-4) ) \ d lambda #

#= int_0^1 \ - (2 - lambda )(1 - lambda) - (1 - lambda)(5 - 5 lambda) - 4 (5 - 4 lambda)(2 - lambda) \ d lambda #

#= int_0^1 \ -22 lambda^2 + 65 lambda - 47 \ dlambda = - 131/6#

  • <1,0,1> to <3,1,2>

#bb r_3 = langle 3,1,2 rangle + lambda langle 2,1,1 rangle, qquad d bb r_3 = langle 2,1,1 rangle d lambda #

#I_3 = int_0^1 \ (((3 + 2 lambda)(1+ lambda)),((1+ lambda)(2 + lambda)),((2 + lambda)(3 + 2 lambda)))\ cdot ((2),(1),(1) ) \ d lambda #

#= int_0^1 d lambda \ 2 (3 + 2 lambda )(1 + lambda) + (1 + lambda)(2 + lambda) + (2 + lambda)(3+ 2 lambda) #

#= int_0^1 \ 7 lambda^2 + 20 lambda + 14 \ dlambda = 79/3#

  • Adding it all up

  • #I_1 + I_2 + I_3 = = 55/2#