# Thread: [SOLVED] Check out my result : line integral

1. ## [SOLVED] Check out my result : line integral

$\displaystyle I=\int_C \frac{dx+dy}{x^2+y^2}$ where $\displaystyle C$ is described by $\displaystyle r(t)=(\cos t, \sin t)$, $\displaystyle 0\leq x \leq 2\pi$.
My attempt : $\displaystyle |r'(t)|=1$ so $\displaystyle I=\int_0^{2\pi} \frac{\cos t dt+\sin tdt}{\cos^2t + \sin^2t}=0$.
I realize that $\displaystyle C$ is a circle of radius $\displaystyle 1$ and that if my result is good then $\displaystyle F(x,y)=\frac{dx+dy}{x^2+y^2}$ is a conservative field. But I'm not sure I've done it right.

2. Originally Posted by arbolis
$\displaystyle I=\int_C \frac{dx+dy}{x^2+y^2}$ where $\displaystyle C$ is described by $\displaystyle r(t)=(\cos t, \sin t)$, $\displaystyle 0\leq x \leq 2\pi$.
My attempt : $\displaystyle |r'(t)|=1$ so $\displaystyle I=\int_0^{2\pi} \frac{\cos t dt+\sin tdt}{\cos^2t + \sin^2t}=0$.
I realize that $\displaystyle C$ is a circle of radius $\displaystyle 1$ and that if my result is good then $\displaystyle F(x,y)=\frac{dx+dy}{x^2+y^2}$ is a conservative field. But I'm not sure I've done it right.
you didn't do it right.

here we have $\displaystyle x = \cos t \implies \boxed{dx = - \sin t~dt}$ and

$\displaystyle y = \sin t \implies \boxed{dy = \cos t~dt}$

(the answer is still 0 though )

3. Thanks for the correction. I do still get 0 as you pointed out.

4. Originally Posted by arbolis
Thanks for the correction. I do still get 0 as you pointed out.
indeed. by now you should have it memorized that integrating sine or cosine over any multiple of their period yields zero. so for instances, $\displaystyle \int_0^{2 \pi} \sin t~dt = \int_\pi^{3 \pi} \sin t~dt = \int_{\pi /2}^{5 \pi / 2} \cos t~dt = \int_\pi^{9 \pi} \sin t~dt$ $\displaystyle = \int_a^{a + 2k \pi} \sin t ~dt = \int_k^{2n \pi} \cos t~dt = 0$ etc etc etc