Vrftsjtryk

I'm starting on double and line integrals, and I'm stuck at this question.
It asks of me to calculate the following integral without using Green's theorem. Usually with Green's theorem I use polar coordinates easily, but I can't seem to make the same substitution here.

$int_{L}{(2x-y)dx+(x-y)dy}$

$L = {(x,y): x^{2}+y^{2}=2y, xgeq0}cup{(x,y):x^{2}+y^{2}=4, xleq0,ygeq0} $

The curve $L$ is oriented from the point $(0,0)$, that is counter-clockwise.

I am not sure which subsitution to use. I always used Greens theorem so far. How would I go about solving this without it? I thought of using some parameter $t$ but couldn't quite specify its range. Thank you.

You have two curves.

first part.

$x = 2sin tcos t = sin 2t\
y = 2sin^2 t = 1-cos 2t$

$int_limits{0}^{frac {pi}{2}} (2sin 2t + 1 - cos 2t )(2cos 2t)+(sin 2t +1-cos 2t)(2sin 2t) dt$

and the second part

$x = 2cos t\
y = 2sin t$

$int_limits{frac {pi}{2}}^{pi} (4cos t - 2sin t)(-2sin t)+(2cos t - sin t)(2cos t) dt$

If you wanted to use greens.

Then the you would add the straight line from $(-2,0)$ to $(0,0)$ to close the contour.

Then

$int_{C_1} P(x,y) dx + Q(x,y) dy + int_{C_2} P(x,y) dx + Q(x,y) dy = iint frac {partial P}{partial y} - frac{partial Q}{partial x} dA\
int_{C_1} P(x,y) dx + Q(x,y) dy = iint frac {partial P}{partial y} - frac{partial Q}{partial x} dA- int_{C_2} P(x,y) dx + Q(x,y) dy$