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

  • $\begingroup$ what do you mean by negative orientation, Is it anticlockwise ? $\endgroup$
    – Learning
    Apr 27, 2020 at 17:14
  • $\begingroup$ No, it is clockwise. $\endgroup$
    – user9867
    Apr 27, 2020 at 17:17
  • $\begingroup$ N what does this tell " Γ is the boundary curve formed with the AB arch of the circle of C centre"? $\endgroup$
    – Learning
    Apr 27, 2020 at 17:20
  • $\begingroup$ Points A and B are joined with a circumference whose centre is point C. $\endgroup$
    – user9867
    Apr 27, 2020 at 17:22
  • 1
    $\begingroup$ 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 $Q=x-y \arctan y$. $\endgroup$
    – memerson
    Apr 27, 2020 at 17:27

1 Answer 1



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


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