Use the gradient theorem. Since F is the gradient of the scalar function f(x, y) = sin(x - 8y), the line integral of F along any path C is equal to
[tex]\displaystyle \int_C \vec F \cdot d\vec r = \int_{(a,b)}^{(c,d)} \nabla f(x, y) \cdot d\vec r = f(c,d) - f(a,b)[/tex]
where (a, b) and (c, d) are any distinct points. (They have to be distinct, otherwise the path would be a closed loop or a single point.)
(a) You just need to pick two points (a, b) and (c, d) from the (x, y)-plane such that
f(c, d) - f(a, b) = 0
where the path C₁ starts at (a, b) and ends at (c, d).
Let (a, b) be the origin, (0, 0). Then we want to find c and d such that
sin(c - 8d) - sin(0 - 8•0) = sin(c - 8d) = 0
⇒ c - 8d = arcsin(0) + 2nπ or c - 8d = π - arcsin(0) + 2nπ
(where n is any integer)
⇒ c - 8d = 2nπ or c - 8d = (2n - 1)π
If we also fix d = 0, then we have infinitely many choices for c, other than c = 0 since that would make (a, b) and (c, d) the same point. So let's pick n = 1 for the first solution set, so that c = 2π.
Our choice for C₁ can be any curve we wish, but let's just take the simplest one: the line segment joining (0, 0) and (2π, 0).
(b) Same as part (a), but this time we want
f(c, d) - f(a, b) = 0
We again take (a, b) to be the origin. Then we want c and d such that
sin(c - 8d) = 1
⇒ c - 8d = arcsin(1) + 2nπ or c - 8d = π - arcsin(1) + 2nπ
⇒ c - 8d = π/2 + 2nπ
Again, we're free to choose d = 0. Then if n = 0, we get c = π/2, and the path C₂ could be the line segment joining (0, 0) and (π/2, 0).
