Question

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?

Answer 1

`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`