Hi, above is the identity of Greens Theorem. Say $$F_1=3x-y$$ and $$F_2=x$$ and $R$ is the region bounded by $y=1+x$ and $y=(x-1)^2$ for $0 \leq x \leq 3$.
I am trying to verify Green's Theorem for this $R$ by evaluating both sides of the inequality. Doing the left hand side, I do the double integral and get 9 (correct me if I am wrong!). However, I am unsure how to evaluate the right hand side. I've tried a few things but did not get the same answer as the left hand side. I am also not sure what the region C is, like I know that the region R is bounded by C, but am unsure how to proceed.
Any help is appreciated! Thanks.
That's your region $R$.
The upper segment of the boundary $C$ is the line $y = 1 + x, \; 0 \leq x \leq 3$. Let's call it $C_1$. On $C_1$: $dy = dx$, $F_1 = 3x - y = 3 x - (1 + x) = 2x -1$, $F_2 = x$.
$$\int_{C_1} (F_1 \, dx + F_2 \, dy) = \int\limits_0^3 (F_1 + F_2) \, dx = \int\limits_0^3 (3x-1) \, dx$$
The lower segment of $C$ is the line $y = (x-1)^2, \; 0 \leq x \leq 3$. Let's call it $C_2$. On $C_2$: $dy = 2 x dx$, $F_1 = 3 x - (x-1)^2$, $F_2 = x$.
$$ \int\limits_{C_2} (F_1 \, dx + F_2 \, dy) = \int\limits_0^3 (F_1 + 2 x F_2) \, dx = \int\limits_0^3 (3 x - (x-1)^2 + 2 x^2 ) \, dx. $$
The last step is to sum them up
$$ \oint_C (F_1 \, dx + F_2 \, dy) = \int\limits_{C_1} (F_1 \, dx + F_2 \, dy) + \int\limits_{C_2} (F_1 \, dx + F_2 \, dy) $$