green's theoremmultivariable-calculus
Consider the planar figure described by the equation $x^4=x^2-y^2$, (which is called the Lemniscate of Gerono). It may be conveniently represented by the parametrization $$x(t)=\sin t,\; y(t)=\sin t \cos t,\; 0\le t\le 2\pi $$
Use Green's theorem to find the area of the closed loop in the region $x\ge0$.
(Suggestion: Choose $P(x,y)=0, Q(x,y)=x$)
I don't understand the parametrization and how to set up the integral.
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: $$\int_0^{2\pi} (\cos(t) - 1)(1 - \cos(t)) dt = - \int_0^{2\pi} 1 - 2\cos(t) + \cos^2(t) dt = -3\pi. $$
Let $A$ denote the area of one of the five triangular "points" and $B$ the area of the central pentagon. Your Green's theorem integral double-counts $B$, so its value is $5A + 2B$, whereas you want to find $5A + B$.
Let $t_{0}$ denote the first root of $x(t)$ larger than $\pi$, i.e., the time at which the curve first crosses the negative $y$-axis after reaching the very top of the curve, and let $C'$ denote the quasi-triangular curve obtained by following $C$ from $\pi$ to $t_{0}$, the using the vertical segment to close up (shaded). (Very roughly, $t_{0}$ is in the vicinity of $1.575\pi$.)
Twice the "Green's theorem integral" over $C'$ is equal to $3A + B$.
Since $$ B = (15A + 6B) - (15A + 5B) = 3\oint_{C} - 10\oint_{C'}; $$ the desired area is $$ 5A + B = \oint_{C} - B = 10\oint_{C'} - 2\oint_{C}. $$
Best Answer
With $Q_x-P_y=1$ we have, via green's theorem, that (with the suggested choice)
$$A = \iint\limits_D 1 \,dA = \oint\limits_{C}{{x\,dy}}$$
Your loop is drawn out by $0\le t\le\pi$.
$$A = \oint\limits_{C}{{x\,dy}} = \int\limits_{0}^{\pi}{\sin (t)\,d(\sin(t)\cos(t))}$$
$$=\int\limits_{0}^{\pi}{\sin (t)(\cos^2(t)-\sin^2(t))}\ dt$$
Edit: The orientation of the provided trace is clockwise in this case. So you need to negate the final answer. If it were counter-clockwise then it would not be necessary.