$w(x,y)=\frac{x^2+3y^2+2xy}{3 (x^2+y^2)^\frac{4}{3}} \ dx - \frac{3x^2+y^2+2xy}{3 (x^2+y^2)^\frac{4}{3}} \ dy$ , calculate $\int_{+\gamma} w $

Question

$\gamma$ is the curve of this equation: $$\rho=e^{-\theta} \qquad \theta \in [0,+\infty)$$

It is oriented in the growing $\theta$

$$w(x)= \sum_{i=1}^n a_i(x) \ dx_i $$ $$\int_{+\gamma} w=\sum_{i=1}^n \int_a^ba_i(\gamma(t)) \ \gamma'(t)$$

I think that $a=0$ and $b=+\infty$

But, I don't know how finding the parametric equation of $\gamma$
How can I find that parametric equation?

Thanks!

Answer 1

I take it as $\rho=e^{-\theta}$ is $\gamma$ written in polar form with $\rho$ being the radius and $\theta$ the angle.

Then $$ \left\{ \begin{aligned} x(\theta)&=\rho \cos \theta=e^{-\theta}\cos\theta\\ y(\theta)&=\rho \sin \theta=e^{-\theta}\sin\theta \end{aligned} \right. $$ Inserting this into the definition of curve integral, doing some simplifications, I end up with $$ \int_\gamma w=\int_0^{+\infty}-\frac{2}{3}e^{-\theta/3}(2\cos\theta+\sin\theta)\,d\theta=-1. $$ Just ask if you need more details on some part.