Calculate the line integral of the scalar field over the curve L: $$ \int_L(x+y)\,ds $$ with $L$ the right loop of $r^2=2\cos(2\theta)$
I've been going at it for over 3 hours now. With several people in a discord for math that were trying to help me but we can't seem to get the solution that should be 2sqrt(2).
Can someone help me get to the solution? :)
I tried the following:
x = r.cos(θ) and y = r.sin(θ) and my $ds= \sqrt{r²\:+\:r'²}d\theta =\sqrt{\frac{2}{cos(2\theta)}} .d\theta$
from there I got my boundaries as θ = $\frac{\pi }{4}$ -> r = 0 and as θ = 0 -> r = $\sqrt{2}$
from here I got that the integral should be the following:
$\int _Lr\left(cos\theta +sin\theta \right)\:\cdot \sqrt{\frac{2}{cos\left(2\theta \right)}}\:.d\theta$
Now the issue is that I don't know what I have to fill in for the boudnaries of my integral. And even if this is correct what I'm doing.
To get the right loop we need $-\tfrac\pi4\leq\theta\leq\tfrac\pi4$ (note that many values of $\theta$ are impossible because they make $\cos(2\theta)<0$). So our curve is $$ s(\theta)=(\sqrt{2\cos (2\theta)}\,\cos\theta,\sqrt{2\cos(2\theta)}\,\sin\theta). $$ Then we need $$ \|s'(\theta)\|=\frac{\sqrt2}{\sqrt{\cos2\theta}}. $$ Then, as $r=\sqrt{2\cos(2\theta)}$, \begin{align} \int_L(x+y)\,ds&=\int_{-\tfrac\pi4}^{\tfrac\pi4}\left(\sqrt{2\cos (2\theta)}\cos\theta+\sqrt{2\cos (2\theta)}\sin\theta\right)\,\left(\frac{\sqrt2}{\sqrt{\cos2\theta}}\right)\,d\theta\\ \ \\ &=2\int_{-\tfrac\pi4}^{\tfrac\pi4}(\cos\theta+\sin\theta)\,d\theta\\ \ \\ &=2\sqrt2. \end{align}