Hmmm...I do understand your parametrization of the second path but it could be
$$x=0\,\,,\,\,dx=0\,\,\,,\,\,y=t\,\,,\,dy=dt\,\,,\,t\in[0,2]$$
and we take the path upwards (and then we can change the sign), so
$$\int_0^22t^2\,dt=\left.\frac{2}{3}t^3\right|_0^2=\frac{16}{3}$$
Thus, the value of the integral is
$$\frac{8}{3}-\frac{16}{3}=-\frac{8}{3}$$
I think that when you got into the integral you forgot to take $\,y^2\,$ , and it should be:
$$\int_0^12\left[2(1-t)\right]^2(-2)dt=-16\int_0^1(1-t)^2dt=\left.\frac{16}{3}(1-t)^3\right|_0^1=-\frac{16}{3}$$
which is what I got above (with the sign changed, of course)
You say that $\textbf{r}(1)=(a,b)$, so $a^2+b^2=100$. But $(a,b)$ is not distance $10$ from the origin, rather distance $10$ from $(2,2)$. So what you know is that $\|\langle a-2,b-2\rangle\|=10$, i.e., $(a-2)^2+(b-2)^2=100$. I think you can continue with the optimization the way you have.
There is an easier way though. Note that $\nabla f=\langle 8,6\rangle$ is a constant vector field. The line integral $\int_C \nabla f \cdot d\textbf{r}$ is by definition $\int_C \nabla f \cdot \textbf{n} \ ds$, where $\textbf{n}$ is a unit tangent vector to the curve $C$ at every point. Since $C$ is a straight line, $\textbf{n}$ is constant along the curve. So $\nabla f \cdot \textbf{n}$ is constant as well, with $\nabla f \cdot \textbf{n}=\|\langle 8,6\rangle\|\|\textbf{n}\| \cos \theta = 10\cos \theta$, where $\theta$ is the angle between $\nabla f=\langle 8,6\rangle$ and $\textbf{n}$. Thus $$\int_C\nabla f \cdot d\textbf{r}=\int_C \nabla f \cdot \textbf{n} \ ds=10\cos \theta \int_C 1 \ ds = 10 \cos \theta ~\text{length}(C)=100\cos \theta.$$
This is maximized when $\cos\theta=1$, i.e., $\theta=0$. In this case, $\textbf{n}$ is a positive scalar multiple of $\nabla f= \langle 8,6\rangle$, so $C$ goes in the same direction as $\langle 8,6\rangle$. As it happens, $\langle 8,6\rangle$ is already a vector of length $10$, so you know the endpoint of $C$ must be $\langle 2,2\rangle + \langle 8,6 \rangle = \langle 10,8\rangle$.
Best Answer
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 .