Question:
Suppose $F(x,y)=2sin(x/2)sin(y/2)i - 2cos(x/2)cos(y/2)j$. C is the curve from $P$ to $Q$, where $P(-3\pi/2,3\pi/2),Q(-3\pi/2,-3\pi/2),R(3\pi/2,3\pi/2),S(3\pi/2,-3\pi/2)$.The curves $PR$ and $SQ$ are trigonometric function of period 2\pi and amplitude 1.
My attempt:
integrate C $Fdr$ = integrate C $F(x(t),y(t))r'(t)dt$, so I split the question into 3 parts:
1. integrate t from -3pi/2 to 3pi/2 $<2sin(t/2)sin(cos(t)/2), -2cos(t/2)cos(cos(t)/2)>⋅<1,-sin(t)>dt$
2. integrate t from 0 to 1
$<2sin(3pi/4)sin((3pi/2-3pi*t)/2), -2cos(3pi/4)cos((3pi/2-3pi*t)/2)>⋅<0,-3pi/2>dt$
3. integrate t from 3pi/2 to -3pi/2
$<2sin(t/2)sin(sin(t)/2), -2cos(t/2)cos(sin(t)/2)>⋅<1, cos(t)>dt$
Are my steps correct? But they are even too complicated that I cannot use calculator to solve them.
For $F(x,y)=A(x,y)\vec{i}+B(x,y)\vec{j}$, if $\displaystyle \frac{dA}{dy}=\frac{dB}{dx}$, then the vector field is conservative and has an "antiderivative".
Consider the potential function $F^*(x,y)=-4\cos(\frac{x}{2})\sin(\frac{y}{2})$.
The derivative (or vector field) corresponding to $F^*$ is F, as we can see through differentiation.
Now, we can just take $F^*(Q)-F^*(P)$.