Given the points $A(2,0), B(1,-1), C(1,0)$ and $D(0,-1)$ in $\mathbb{R}^2$, using Green's theorem I have to calculate the following integral:
$$\int_{\Gamma}(x^4 -x^3e^x-y)dx+(x-y \arctan y)dy$$
Where $\Gamma$ is the boundary curve formed with the AB arch of the circle of C centre and the segments BD, DO and OA, where O is the origin of coordinates. All of this with negative orientation.
First of all, I've stated that $F_1 = P$ and $F_2 = Q$, and then:
$$\frac{\partial P}{\partial y} = -\arctan y - \frac{y}{1+y^2}$$ $$\frac{\partial Q}{\partial x} = 4x^3-3xe^x-x^3e^x$$
But then, I don't know how to state the $\Gamma$ set. I've written that:
$$\Gamma = [ (x,y) | -1 \le y \le 0; 0 \le x \le 1 ]\cup [ (r,\theta) | 0 \le \theta \ \pi /2; 0 \le r \le 1 ] $$
But I don't know if it's correct. Then I've solved the P integral and the Q integral is impossible to solve due to $sin$ and $cos$ expressions.
In the exercise solution, it says:
$$\frac{\partial P}{\partial y} = -1$$ $$\frac{\partial Q}{\partial x} = 1$$
Why? I don't understand it.
Thanks in advance
Hint:
Note that $P$ and $Q$ are continuously differentiable in the region $R$(say) bounded by closed curve $\Gamma$ which consists arc $AB$ of a circle centered at point $C$, line segments $BD, DO,OA$. So we apply Green's Theorem:$$\int_{\Gamma(aniclockwise)}Pdx+Qdy=\iint_{R}\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}.$$
Here, $P= x^4 -x^3e^x-y$, $\frac{\partial P}{\partial y}=-1$ and $Q=x-y \arctan y$, $\frac{\partial Q}{\partial x}=1$. Therefore, the line integral reduced as: $$\color{Red}{-}2\iint_{R}dxdy=\overset{clockwise}{-}2\int_{y=-1}^{0}\int_{x=0}^{1-\sqrt{1-y^2}}dxdy.$$