## Question

Evaluate the line integral int_C[(4y+5)dx+2xzdy+(yz-x)dz] along the following paths C:

1. C is defined parametrically by x=2t^2, y=6t, z=t^3 from t=0 to t=1.
2. C consists of straight lines from (0,0,0) to (0,1,0) then to (0,1,1) and then to (6,1,1).
3. C is given by the straight line joining (0,0,0) to (1,2,6).

int_C[(4y+5)dx+2xzdy+(yz-x)dz].

1. x=2t^2, y=6t, z=t^3, 0<=t<=1.

int_C[(4y+5)dx+2xzdy+(yz-x)dz]=

=int_0^1[(4*6t+5)*(2t^2)'+2*2t^2*t^3*(6t)'+(6t*t^3-2t^2)*(t^3)']dt=

=int_0^1[(24t+5)*(4t)+24t^5+(6t^4-2t^2)*3t^2]dt=int_0^1[18t^6+24t^5-6t^4+96t^2+20t]dt=

=(18/7t^7+24/6t^6-6/5t^5+96/3t^3+20/2t^2)|_0^1=18/7+24/6-6/5+96/3+20/2=1658/35.

2. C consists of 3 paths:

1. Line from (0,0,0) to (0,1,0).

Parameterization is given as x=0, y=t, z=0, 0<=t<=1.

int_C[(4y+5)dx+2xzdy+(yz-x)dz]=

=int_0^1[(4*t+5)*(0)'+2*0*0*(t)'+(t*0-0)*(0)']dt=0.

2. Line from (0,1,0) to (0,1,1).

Parameterization is given as x=0, y=1, z=t, 0<=t<=1.

int_c[(4y+5)dx+2xzdy+(yz-x)dz]=

=int_0^1[(4*1+5)*(0)'+2*0*t*(1)'+(1*t-0)*(t)']dt=

=int_0^1tdt=t^2/2|_0^1=1/2.

3. Line from (0,1,1) to (6,1,1).

Parameterization is given as x=6t, y=1, z=1, 0<=t<=1.

int_c[(4y+5)dx+2xzdy+(yz-x)dz]=

=int_0^1[(4*1+5)*(6t)'+2*6t*1*(1)'+(1*1-6t)*(1)']dt=

=54int_0^1dt=54.

Required integral is sum of 3 line integrals: 0+1/2+54=54.5.

3. Line from (0,0,0) to (1,2,6).

Parameterization is given as x=t, y=2t, z=6t, 0<=t<=1.

int_c[(4y+5)dx+2xzdy+(yz-x)dz]=

=int_0^1[(4*2t+5)*(t)'+2*t*6t*(2t)'+(2t*6t-t)*(6t)']dt=

=int_0^1[(4*2t+5)+2*t*6t*2+(2t*6t-t)*6]dt=int_0^1[96t^2+2t+5]dt=

=(96/3t^3+2/2t^2+5t)|_0^1=96/3+1+5=38.