Question

F = (y^2 + 1)i + (x^2 + 4)j and
C : the triangle bounded by y = 0, x + y = 2, y = x.
Use Green’s Theorem to find counterclockwise circulation

Answer 1

so i think it is \[\int\limits_{0}^{2}\int\limits_{0}^{x-2} Mdy-Ndx\]

Answer 2

which is \[\int\limits_{0}^{2}\int\limits_{0}^{2-x} 2x-2y dxdy\]

Answer 3

Well
\[\oint\limits_C \vec{F} \cdot d \vec{l}=\oint\limits_C M dx + N dy =\iint\limits_D \left( \frac{\partial N}{\partial x} -\frac{\partial M}{\partial y}\right)dxdy\]
\[D: \left\{ (x,y)| y \le x \le 2-y; 0 \le y \le 1 \right\}\]

Answer 4

i understand how to set it up solving i am messin up

Answer 5

Setting up is the hard part :P Where is the evaluation giving you trouble?

Answer 6

i get 2x-2y dx dy

Answer 7

then i get X^2-2xy

Answer 8

then i plug in i get (2-x)^2 - 2(2-xy)

Answer 9

Once you integrate with x you shouldn't get a function of x back out. It should ONLY be a function of y.

Answer 10

i know but what am i doing wrong