Calculate the line integral: $\vec{F}(x,y)=(x^{2}+y^{2})i+(x^{2}-y^{2})j$

Question

Calculate the line integral of, $$\vec{F}(x,y)=(x^{2}+y^{2})i+(x^{2}-y^{2})j$$ along the curve described
a) by the equation: $y=1-|1-x|$ from $(0,0)$ at $(2,0)$

b) the closed curve described by the equations $y=1-|1-x|$ and $y=0$ in a counterclockwise direction

I need to expose the exercise

Answer 1

(a) Parametrizing the given curve $\mathscr{C}: x=t, y=t$ $ 0\leq t \leq 1$ $$x=t,y=-t+2, 1 \leq t \leq 2$$

$\implies$ $\int_{\mathscr{C}}\vec{F}.d\vec{R}=\int_{0}^{1}2t^2dt +\int_{1}^{2} 2(2-t)^2 dt$

2. Use green's theorem.

Answer 2

For a) Firts, parametrize the curve. The next parametrization works: \begin{equation} \gamma(x)= \left\{ \begin{array}{ccc} (x,x) & \text{if} & x\in[0,1]\\ (x-1)(2,0)+(2-x)(1,1) & \text{if} & x\in [1,2] \end{array} \right. \end{equation} Then, split the integral in two parts $$\displaystyle\int_0^1 F(\gamma)\gamma ' + \displaystyle\int_1^2 F(\gamma)\gamma '$$Only substitute the values to finalize the exercise.

For b) First, recall that if $F:\mathbb{R}^2\to\mathbb{R}^2$ is a field and $\gamma:[a,b]\to\mathbb{R}^2$ then the line integral of $F$ over the curve $\gamma$ is $\displaystyle\int_a^b F(\gamma)\cdot\gamma'$. You need to parametrize the triangle formed by the curves $f(x)=0$ and $g(x)=1-|1-x|$. You can do it with a piecewise function like this: (note that this parametrization is counterclockwise).

\begin{equation} \gamma(x)= \left\{ \begin{array}{ccc} (x,0) & \text{if} & x\in[0,2]\\ (x-2)(1,1)+(3-x)(2,0) & \text{if} & x\in[2,3]\\ (x-3)(0,0)+(4-x)(1,1) & \text{if} & x\in [3,4] \end{array} \right. \end{equation}

Finally, the field in the exercise is $F(x,y)=(x^2+y^2,x^2-y^2)$. You can split the line integral in three parts: $$\displaystyle\int_0^2 F(\gamma)\gamma ' +\displaystyle\int_2^3 F(\gamma)\gamma ' +\displaystyle\int_3^4 F(\gamma)\gamma '$$ Can you finish from here? Only substitute the values again.