Suppose $f(x,y)=x^2y^2$ and path defined $\alpha(t)=(\cos t, \sin t)$, $0\leq t\leq 4\pi$
Going to find $\int_{\alpha}f ds$, but not completely sure in the final part.
Ok, first $x(t)=\cos t$ and $y(t)=\sin t$, so $dx/dt=-\sin t$ and $dy/dt=\cos t$
Thus $ds=\sqrt{\sin^2 t + \cos^2 t}dt=1dt$. So far so good.
Next $\int_{\alpha}fds=\int_0^{4\pi}\cos^2 t \sin^2 t dt$, and next step is what I am uncertain of - if I simply plug $t$ values in there, I get $0$.
$\alpha(t)$ goes in circle, and makes two full circle. If I plug anything ($\pi/x|x\in\mathbb{N}$), larger than $\pi/4$ I also get $0$.
for $t$ from $0$ to $\pi/4$ - $\int_{\alpha}=1/4$, and since we have $16$ such parts, full $\int=4$
Would someone be as kind, as to help me interpret - what I am to aim for in this case?
It seems to me that I make mistage by "plugging" t into integrand, which actually doesnt make any sense.
With mistake corrected $\int_0^{4\pi} cos^2(t)*sin^2(t)dt=\rvert_0^{4\pi}1/8t - 1/32*sin(4t)=1/2\pi$, which now makes way more sense, because if we to integrate only one "full circle", result is $1/4\pi$