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)`.

Answer

`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`.