I am lost here: $C = x^2 + y^2 = 4$ from $(0,2)$ to $(-2, 0)$. Calculate $ \ \int_c y^2 ds \ \ $ and give reasons the sign is correct.
It's obviously the circular arc going counterclockwise from (0,2) to (-2,0), how do you figure out whether $dx$ and $dy$ are positive or negative?
My attempt is:
$$r(t) = < 0, \ 2 sin(t)>,\ \frac{\pi}{2} \le t \le \pi$$ $$\frac{dr}{dt} = <0, 2 cos(t)>$$ $$ ds = \sqrt {2cos^2(t)} \ dt$$ $$\int_C \ y^2 \ ds = \int_{\frac{\pi}{2}}^{\pi} \ 2 sin^2(t) \ \sqrt{2cos^2(t)} \ dt$$ I'm not sure if that's correct or where to go from here. Any help is greatly appreciated.
As you've found now, the arclength integration on the quarter-circle (using what is essentially angle-parametrization) in the second quadrant is
$$ \int_{\pi / 2}^{\pi} \ \ (2 \ \sin \ t)^2 \ \ \sqrt{(-2 \sin \ t)^2 \ + \ (2 \cos \ t)^2} \ \ dt \ \ = \ \ \int_{\pi / 2}^{\pi} \ \ 4 \ \sin^2 t \ \cdot \ 2 \ \ dt \ \ $$
$$ = \ \ 8 \ \int_{\pi / 2}^{\pi} \ \ \frac{1}{2} ( \ 1 \ - \ \cos \ 2t \ )\ \ dt \ \ = \ \ ( \ 4 \ t \ - \ 2 \ \sin \ 2 t \ ) \ \vert_{\pi / 2}^{\pi} $$
$$ = \ \ ( \ 4 \ \pi \ - \ 2 \ \sin \ 2 \ \pi \ ) \ - \ ( \ 4 \ \cdot \frac{\pi}{2} \ - \ 2 \ \sin \ 2 \cdot \frac{\pi}{2} \ ) $$
$$ = \ 4 \ \pi \ - \ 0 \ - \ 2 \ \pi \ + \ 0 \ = \ 2 \ \pi \ \ . $$
Here is a graph of the tangent vector $ \ \mathbf{r}(t) \ $ following the circular arc. Both $ \ dx \ $ and $ \ dy \ $ are negative in the second quadrant $ ^* $, since the $ \ x \ $ and $ \ y \ $ coordinates of points along the circle are decreasing as the vector "moves" in the specified direction. Nonetheless, $ \ ds \ $ , the infinitesimal arclength element is always positive; since $ \ y^2 \ $ is non-negative, the result of the integration should have a positive value.
$ ^* $ The fact that $ \ dy \ $ is negative on this arc is the reason ellya obtained a negative value from integrating $ \ \int_C \ y^2 $ dy .