Question

F = (x2 y2)i (x - y)j; c is the rectangle with vertices at (0, 0), (3, 0), (3, 9), and (0, 9)

Answer 1

I assume
`C` is oriented counterclockwise. If
`C` is the closed rectangle, then by Green's theorem we have

`\displaystyle \int_C F\cdot dr = \int_0^3 \int_0^9 (\partial (x-y))/(\partial x) - (\partial (x^2+y^2))/(\partial y) \, dy \, dx \\\\ ~~~~~~~~~~~~ = \int_0^3 \int_0^9 (1 - 2y) \, dy \, dx = 3 \int_0^9 (1-2y)\, dy = \boxed{-216}`

It seems unlikely, but if you actually are omitting the integral along the line segment joining (0, 9) and (0, 0), let
`x=0` and
`y=9-t` with
`0\le t\le9`. Then the integral along this line segment would be

`\displaystyle \int_((0,9)\to(0,0)) F\cdot dr = \int_0^9 \bigg((0^2 + (9-t)^2) i + (0 - (9-t)) j\bigg) \cdot (0i - j) \, dt \\\\ ~~~~~~~~  = \int_0^9 (9-t) \, dt = \frac{81}2`

so that the overall integral would instead have -216 - 81/2 = -513/2.