I'm learning Green's Theorem right now but I still get confused with identifying which one is P and Q. I am also not sure with my bounds of integration. I wish to validate my solution with the problem:
Let $C_1$ be the line segment from (0,1) to (1,0), $C_2$ be the portion of the parabola $x = (y+1)^2$ from (1,0) to (4,1) and $C_3$ be the line segment from (4,1) to (0,1).
Use Green's Theorem to evaluate
\begin{align*} \oint_{C_1\cup C_2\cup C_3}[2x^2+\tanh(x^2+1)]\,dx + [xe^{2y^3+9y^2}-\sin(4y^3)]dy \end{align*}
My approach:
Let $Q = [xe^{2y^3+9y^2}-\sin(4y^3)] dy$, $P = [2x^2+\tanh(x^2+1)] dx$
Then, $\dfrac{\partial Q}{\partial x} = e^{2y^3+9y^2}$, $\dfrac{\partial P}{\partial y} = 0$
\begin{align*} \oint_{C_1\cup C_2\cup C_3}[2x^2+\tanh(x^2+1)]\,dx + [xe^{2y^3+9y^2}-\sin(4y^3)]\,dy &= \int\int_R\dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y}dA\\ &=\int\int_R(e^{2y^3+9y^2}-0)\,dA\\ &=\int_0^1\int_{1-y}^{3y+1} e^{2y^3+9y^2}\,dxdy\\ &=\int_0^1 e^{2y^3+9y^2}(x)\bigg|_{1-y}^{3y+1}\,dy\\ &=\int_0^1 e^{2y^3+9y^2}(3y+1 - (1-y))\,dy\\ &=\int_0^1 4ye^{2y^3+9y^2}\,dy \end{align*}
My solution is not yet complete, but I want to know if I am on the right direction.
Your approach is correct but you have a mistake in the upper bound of $x$ in your integral. From point $(1, 0)$ to point $(4,1)$, you are on parabola $x = (y+1)^2$.
So the bound of $x$ should be $1-y \leq x \leq (y+1)^2$ and for $y$ it is $0 \leq y \leq 1$.
$ \displaystyle I = \int_0^1\int_{1-y}^{(y+1)^2} e^{2y^3+9y^2}\,dx \ dy$
$ \displaystyle I = \int_0^1 (y^2+3y) \ e^{2y^3+9y^2} \ dy$
Now substituting $2y^3 + 9y^2 = t$ makes the integral straightforward.
Below is a diagram that should explain the bounds.