multivariable-calculus
I want to use Green's theorem for computing the area of the region bounded by the $x$-axis and the arch of the cycloid:
$$ x = t- \sin (t),\;\;\; y = 1 – \cos (t),\;\; 0 \leq t \leq 2\pi $$
So basically, I know the radius of this cycloid is 1.
And to use Green's theorem, I will need to find $Q$ and $P$.
$$\int_C P\,dx + Q\,dy = \iint_D \left(\frac{\partial Q}{\partial x} – \frac{\partial P}{\partial y}\right) dA$$
No, changing the bounds will not help you here. The area is signed (or "oriented"). An easy way to look at this is to form the cross product $dA = r \times v \, dt$, where $v = dr/dt$. You should be able to see that, for $x > 0$, this cross product is in the positive $z$-direction. For $x < 0$, this cross product is in the negative $z$-direction. The two halves have opposite orientations.
By symmetry, both halves have the same area, but since they're oriented in opposite directions, they cancel out, and you end up with zero.
Maybe a picture helps here:
Notice that union of the black, green, red, green, blue and green gives a simply connected region on which the curl vanishes hence the totality of an integral around that region is zero. The same goes for the union of the curves bounding the lower simply connected region. Furthermore, when we add these two integrals together the green and orange cross-cuts cancel just leaving the black CCW curve and the inner CW oriented integrals around the red and blue circles which contain the singular points $(0,0)$ and $(2,0)$ for your vector field. Notice, the integral around the red and blue circles are easy because only the singular term nontrivially contributes.
This is why your teacher wanted CW-oriented circles. I do think you can deform the circles to a point as Muphrid suggested, but, you can also follow my sketch.
Best Answer
You can take $P = y$ and $Q = 0$. Then, $$ \oint\limits_C-ydx = \iint\limits_D dA = \text{Area}(D). $$ Along the $x$-axis, you have $y = 0$, so you only need to compute the integral over the arch of the cycloid. Note that your parametrization of the arch is a clockwise parametrization, so in the following calculation, the answer will be the minus of the area: $$\int_0^{2\pi} (\cos(t) - 1)(1 - \cos(t)) dt = - \int_0^{2\pi} 1 - 2\cos(t) + \cos^2(t) dt = -3\pi. $$