<s>[INST] Evaluate the path integral of $f(x, y) = y$ over the graph of the semicircle $y = \sqrt{1-x^2}, -1 \leq x \leq 1$ Evaluate the path integral of $f(x, y) = y$ over the graph of the semicircle $y = \sqrt{1-x^2}, -1 \leq x \leq 1$
Solution attempt:
$f(x, y) = y$, along $y = \sqrt{1-x^2}, -1 \leq x \leq 1$
$\vec{r}(t) = (t, \sqrt{1-t^2}), -1 \leq t \leq 1$
$\vec{r}'(t) = (1, \frac{-2t}{2\sqrt{1-t^2}}) = (1, \frac{-t}{\sqrt{1-t^2}})$
$||\vec{r}'(t)|| = \sqrt{1+\frac{t^2}{1-t^2}} = \frac{1}{\sqrt{1-t^2}}$
Show $\int f(x, y) dt = \int_{-1}^{1} \sqrt{1-t^2}||\vec{r}'(t)|| dt = \int_{-1}^{1} \sqrt{1-t^2}\frac{1}{\sqrt{1-t^2}} dt = 1+1 = 2$
therefore the value of the required path integral is 2
is this correct? 
 [/INST] You can also parameterise the curve $x=cost$, $y=sint$
$$r'(t) = <-sint,cost>$$
$$||r'(t)|| = 1$$
$$F = sin t$$
$$I = \int_{0}^{\pi} sint dt = -cost|_{0}^{\pi} = 2$$
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