I'm currently revising for a maths module that I am taking as part of my physics degree. I'm taking the exam tomorrow and I'm feeling pretty confident although upon attempting this line integral I drew a blank.
$$ S = \frac{1}{2} \oint_C \vec{r} \times d\vec{r} $$ Evaluate this line integral for a particle moving once anti-clockwise around a unit circle C in the x-y plane defined by the equation $ x^2 + y^2 + z^2 = a^2$.
I guess that my main issue with answering this question is what limits to use since the line both starts and ends at the same point. Usually I'd parametrise a system like this but I'm having no luck.
Whilst this question does seem somewhat basic my tired and frazzled mind is incapable of solving it. Any help would be much appreciated! Thanks, Sean.
Thanks to Marra for your help! For anyone interested this is how I answered it in the end:
$x^2 + y^2 + z^2 = a^2$ is a uniform sphere. It intersects the $x-y$ plane in the form of a uniform circle of radius $a$ and equation $x^2 + y^2 = a^2$. $$ Let:~~~\vec{r} = (a\cos{t},~a\sin{t},~0) ~~~ [From ~t=0~to~t=2\pi]\\\Rightarrow \frac{d\vec{r}}{dt} = (-a\sin{t},~a\cos{t},~0)\\\Rightarrow S=\frac{1}{2}\int_0^{2\pi} \vec{r}\times\frac{d\vec{r}}{dt}dt\\\vec{r}\times\frac{d\vec{r}}{dt} = (0,~0,~a^2)\\\Rightarrow~S=\frac{1}{2}.(0,~0,~a^2).\int_0^{2\pi} dt = \frac{1}{2}.(0,~0,~a^2).2\pi = (0,~0,~\pi a^2) $$ This is the answer is expected since the area of a circle is $\pi r^2$.
Please correct me if I have made any mistakes. Thanks, Sean.