<s>[INST] Using a line integral to find work For the life of me I can't see where I'm going wrong with this.

Given the vector field
$$ F(x, y) = \left[ xy, \frac{2y}{x} \right] $$
a particle travels on the hyperbola 
$$ \frac{x^2}{4} - y^2 = 1 $$
from $ \left(2\sqrt{10}, -3 \right) $ to $ \left(2, 0 \right) $ how much work is done by the force over the path?

My take on this is to solve for x and parametrize $t = y^2 + 1$ such that
$$ x = 2\sqrt{t} $$ $$ y = \sqrt{t-1} $$
$$ t\in\left[10,1\right]$$
for which
$$ \int_C F(r(t)) \cdot r'(t) dt = \int_{10}^1 2\sqrt{t}\sqrt{t-1}\frac{1}{\sqrt{t}} + \frac{2\sqrt{t-1}}{2\sqrt{t}}\frac{1}{2\sqrt{t-1}} $$
$$ = \left. \left(\frac{4}{3}\left(t-1\right)^{\frac{3}{2}} + \sqrt{t} \right) \right|_{10}^1 = -35 - \sqrt{10} $$
would be grateful for any input.
 [/INST] Since the $y$-coordinate of the trajectory is negative, you should parametrize the path as 
$$ x = 2\sqrt{t}, \>\>\> y = -\sqrt{t-1},\>\>\> t\in\left[10,1\right]$$
to allow negative values for the $y$-variable. Then, the work integral becomes
$$ \int_C F(r(t)) \cdot r'(t) dt = \int_{10}^1 \left[2\sqrt{t}(-\sqrt{t-1})\frac{1}{\sqrt{t}} + \frac{2\sqrt{t-1}}{2\sqrt{t}}\frac{1}{2\sqrt{t-1}}\right]dt $$
$$ = \left. \left(-\frac{4}{3}\left(t-1\right)^{\frac{3}{2}} + \sqrt{t} \right) \right|_{10}^1 = 37 - \sqrt{10} $$
</s>