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.


$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.

  • $\begingroup$ Tell your teacher that this is a sick way to define a $1$-chain. $\endgroup$ – Christian Blatter Nov 20 '18 at 19:08

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.


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

  • $\begingroup$ Thanks, I understand everything except how you got the first parts parametric equations. Since the first part of the curve is the right side of a circle of radius 1 and y offset of 1, shouldn't y be $y=1+sin(t)$. And $x=cos(t)$. $\endgroup$ – math101 Nov 20 '18 at 22:32
  • $\begingroup$ You could use that parameterization, it would be for a different range of t. That is fine. I said, $x = r\cos t, y = r\sin t$ plugging into $x^2 + y^2 = 2y$ gives $r = 2\sin t$ And put that back in for $r$ in the equations of $x,y$ $\endgroup$ – Doug M Nov 20 '18 at 22:42
  • $\begingroup$ So that's it! I see now , thanks! $\endgroup$ – math101 Nov 20 '18 at 23:13

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