# Calculate integral using Green's Theorem

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.

• what do you mean by negative orientation, Is it anticlockwise ? Apr 27, 2020 at 17:14
• No, it is clockwise. Apr 27, 2020 at 17:17
• N what does this tell " Γ is the boundary curve formed with the AB arch of the circle of C centre"? Apr 27, 2020 at 17:20
• Points A and B are joined with a circumference whose centre is point C. Apr 27, 2020 at 17:22
• You seem to have made a mistake in defining $P$ and $Q$. You should have that it's $\int P\text{d}x+Q\text{d}y$, and so $P= x^4 -x^3e^x-y$ and $Q=x-y \arctan y$. Apr 27, 2020 at 17:27

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: $$\color{Red}{-}2\iint_{R}dxdy=\overset{clockwise}{-}2\int_{y=-1}^{0}\int_{x=0}^{1-\sqrt{1-y^2}}dxdy.$$