Question

Evaluate the line integral, where C is the given curve. C (x + yz) dx + 2x dy + xyz dz, C consists of line segments (1, 0, 1) to (2, 4, 1) and from (2, 4, 1) to (2, 6, 4)

Answer 1

Parameterize the first line segment $C_1$ by

$\vec r(t)=(1,0,1)(1-t)+(2,4,1)t=(1+t,4t,1)$

and the second line segment $C_2$ by

$\vec s(t)=(2,4,1)(1-t)+(2,6,4)t=(2,4+2t,1+3t)$

both with $0\le t\le1$. Then

$\displaystyle\int_{C_1}(x+yz)\,\mathrm dx+2x\,\mathrm dy+xyz\,\mathrm dz$

$\displaystyle=\int_0^1\bigg((1+5t)\cdot1+2(1+t)\cdot4+(1-t)4t\cdot0\bigg)\,\mathrm dt$

$=\displaystyle\int_0^1(9+13t)\,\mathrm dt=\frac{31}2$

and

$\displaystyle\int_{C_2}(x+yz)\,\mathrm dx+2x\,\mathrm dy+xyz\,\mathrm dz$

$\displaystyle=\int_0^1\bigg((2+(4+2t)(1+3t))\cdot0+2(2)\cdot2+2(4+2t)(1+3t)\cdot3\bigg)\,\mathrm dt$

$\displaystyle=\int_0^1(32+84t+36t^2)\,\mathrm dt=86$

Then

$\displaystyle\int_C(x+yz)\,\mathrm dx+2x\,\mathrm dy+xyz\,\mathrm dz=\frac{31}2+86=\boxed{\frac{203}2}$