A force field in 3-space is given by the formula
$F(x,y,z)=(x+yz,y+xz,x(y+1)+z^2)$. Calculate the work done by F in
moving a particle once around the triangle with vertices $(0,0,0)$,
$(1,1,1)$, $(1,1,0)$ in that order.
So $W=\int \limits_{C} Fdr$ but how do I make my $r(t)$ function? Do I do it part by part as in let A, B, C be the vertices of the triangle respectively with the order they are given. First find the vector equation of AB intersecting with A and B which would be say $r_1(t)= (0,0,0)+(1,1,1)t=(t,t,t)$ right? But this is just a guess. Even if this is right, what would the limits even be.
I missed one lecture on this module so I am unsure. Please help.
Best Answer
In a vector setting $$ W=\int F\cdot\mathrm{d}r $$ We can parametrize the first side of the triangle by $r=(t,t,t)$ for $t$ from $0$ to $1$: $$ \begin{align} W_1 &=\int_0^1(t+t^2,t+t^2,t+2t^2)\cdot(1,1,1)\,\mathrm{d}t\\ &=\int_0^1(3t+4t^2)\,\mathrm{d}t\\ &=\frac32+\frac43\\ &=\frac{17}6 \end{align} $$ We can parametrize the second side of the triangle by $r=(1,1,1-t)$ for $t$ from $0$ to $1$: $$ \begin{align} W_2 &=\int_0^1(2-t,2-t,2+(1-t)^2)\cdot(0,0,-1)\,\mathrm{d}t\\ &=\int_0^1(-2-(1-t)^2)\,\mathrm{d}t\\ &=-2-\frac13\\ &=-\frac73 \end{align} $$ The third side is similar.