Helping understand line integral $\int_{K,+}{(x+y)}dx+(y-x)dy$

Question

I have a huge problem with understanding line integrals and would be much obliged for your help! We have: $$\int_{K,+}{(x+y)}dx+(y-x)dy$$ and the following parameterization: $$K:x=a\cdot\alpha\cdot\cos\alpha$$ $$ y=a\cdot\alpha\cdot\sin\alpha$$ where $a$ is a positive constant and $\alpha\in[0,2\pi]$. First I would like to understand the notation $\int_{K,+}{(x+y)}dx+(y-x)dy$. Is it equal to $\int_{K,+}{(x+y)}dx+\int_{K,+}(y-x)dy$? How should I calculate it?

Answer 1

The notation corresponds to a line integral of a vector field $$\int_{\vec{\gamma}}\vec{f}(\vec{r})\cdot\mathrm{d}\vec{r} = \int_{\vec{\gamma}}\vec{f}(x,y)\cdot\mathrm{d}\vec{r} = \int_{\vec{\gamma}}f_x(x,y)\mathrm{d}x+f_y(x,y)\mathrm{d}y$$ since $\vec{r} = x\vec{i}+y\vec{j}$ and $\vec{f}(x,y)= f_x(x,y)\vec{i}+f_y(x,y)\vec{j}$

It can be solved, by rewriting it, according to the parametrisation of the curve $\vec\gamma: [t_1,t_2] \rightarrow \mathbb{R}^2$: $$\int_{\vec\gamma}\vec f(\vec r)\cdot\mathrm{d}\vec r = \int_{t_1}^{t_2}\vec f(\vec \gamma(t))\cdot \vec \gamma'(t)\mathrm{d}t $$ which is a classical integral over $\mathbb{R}$.

In your case, $\alpha$ is the parameter, then the integration interval will be $[\alpha _1,\alpha _2] = [0,2\pi]$, while the derivative of the path is: $$\vec\gamma'(\alpha) = x'(\alpha)\vec i +y'(\alpha)\vec j$$ where $$ \begin{cases} x'(\alpha)= a\cos\alpha -a\alpha \sin\alpha \\ y'(\alpha)= a\sin\alpha +a\alpha \cos\alpha \end{cases}$$ The function to be integrated is also rather simple: $$f(x,y) = (x+y)\vec i + (y-x)\vec j$$ Eventually, the overall integral: $$ \int_{\alpha_1}^{\alpha_2}\vec f(\vec \gamma(\alpha))\cdot \vec \gamma'(\alpha)\mathrm{d}\alpha = \int_{\alpha_1}^{\alpha_2} \{\, [x(\alpha)+y(\alpha)]x'(\alpha) + [y(\alpha)-x(\alpha)]y'(\alpha) \,\} \mathrm{d}\alpha = \int_{0}^{2\pi} (a^2\alpha\cos^2(\alpha) - a^2\alpha^2\sin\alpha\cos\alpha + a^2\alpha\sin\alpha\cos\alpha - a^2\alpha^2\sin^2(\alpha)) \mathrm{d}\alpha + \int_{0}^{2\pi} (a^2\alpha\sin^2(\alpha) + a^2\alpha^2\sin\alpha\cos\alpha -a^2\alpha\sin\alpha\cos\alpha - a^2\alpha^2\cos^2(\alpha))\mathrm{d}\alpha = \int_{0}^{2\pi} (a^2\alpha - a^2\alpha^2)\mathrm{d}\alpha $$

Answer 2

All right i think i got it more or less.$$\int_{K,+}{(x+y)}dx+(y-x)dy=\int_{0}^{2\pi}(a\cdot\alpha(\cos\alpha+\sin\alpha)a(\cos\alpha-\alpha\sin\alpha)+\alpha\cdot a\cdot(\sin\alpha-\cos\alpha)a(\sin\alpha+\alpha\cos\alpha))d\alpha=...$$ Now i am wondering how we even look for parametrisation. For instance we have : $$\int_{K,+}(2x+y)dx+(x-2y)dy$$ $$K=\{(x,y):x^4+y^3=1;x\le0\le y\}$$ where it's beggining is $(0,1)$ and end $(-1,0)$ how can i find proper parametrisation for it? EDIT: $$x=-(\sin\alpha)^{1/2}$$ $$y=cos(\alpha)^{2/3} $$ and $\alpha\in[0,pi/2]$ seems to be fine?