Line integral multiple path points

Question

What is the work done for a particle that moves from $$(0,0,0)\ \text{ to }\ (0,2,1) \ \text{ to } \ (1,2,1) \ \text{ to }\ (1,0,0) \ \text{ back } \ \text{ to } \ (0,0,0)$$The force is $$F=z^2i+2xyj+4y^2k$$Is the approach to find them all separately and add the works up? What I have done so far is found the work from the origin to the first point but from point $2$ to point $3$ I'm getting $F(r(t))=0$.

Answer 1

$\vec F=(z^2,2xy,4y^2)$

$A(0,0,0)\ \text{ to }\ B(0,2,1) \ \text{ to } \ C(1,2,1) \ \text{ to }\ D(1,0,0) \ \text{ back to} \ A(0,0,0)$

$\gamma_1: r_1(t):=tB+A=(0,\,2t,\,t)\;,\;\;0\le t\le 1$

$\vec F(r_1(t))=(t^2,0,16t^2);\;r_1'(t)=(0,\,2,\,1)$

$\vec F(r_1(t))\cdot r_1'(t)=16t^2$

$W_1=\int_{\gamma_1}\vec F\cdot d\vec r_1=\int_0^1 F(r_1(t))\cdot r'(t)dt=\int_0^1 16t^2\,dt=\dfrac{16}{3}$

in a similar way we get $W_2=\dfrac{139}{3}$ from $B$ to $C$

$W_3=1$ from $C$ to $D$ and $W_4=0$ from $D$ to $A$

the sum is $W=\dfrac{158}{3}$