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, x\geq0\}\cup\{(x,y):x^{2}+y^{2}=4, x\leq0,y\geq0\} $
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.
$x = 2\sin t\cos t = \sin 2t\\ y = 2\sin^2 t = 1-\cos 2t$
$\int_\limits{0}^{\frac {\pi}{2}} (2\sin 2t + 1 - \cos 2t )(2\cos 2t)+(\sin 2t +1-\cos 2t)(2\sin 2t) \ dt$
and the second part
$x = 2\cos t\\ y = 2\sin t$
$\int_\limits{\frac {\pi}{2}}^{\pi} (4\cos t - 2\sin t)(-2\sin t)+(2\cos t - \sin t)(2\cos t) \ dt$
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
$\int_{C_1} P(x,y)\ dx + Q(x,y)\ dy + \int_{C_2} P(x,y)\ dx + Q(x,y)\ dy = \iint \frac {\partial P}{\partial y} - \frac{\partial Q}{\partial x}\ dA\\ \int_{C_1} P(x,y)\ dx + Q(x,y)\ dy = \iint \frac {\partial P}{\partial y} - \frac{\partial Q}{\partial x}\ dA- \int_{C_2} P(x,y)\ dx + Q(x,y)\ dy$