24c01f…: “From a violet bay”

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Consider an analemma:

The left loop of this curve is drawn in a counterclockwise fashion by $\vec {a}(t) = \vector {\sin (t), \sin (2t)}$ as $t$ runs from $\pi$ to $2\pi$ . Let $R$ be the region enclosed by this curve.

Set $\vec {F}(x,y) = \vector {0,x}$ and use Green’s theorem to set-up, but do not evaluate, a line integral computing the area of $R$ . $\iint _R\d A = \int _{\answer {\pi }}^{\answer {2\pi }} \answer {2\sin (t)\cos (2t)} \d t$
Set $\vec {G}(x,y) = \vector {-y,0}$ and use Green’s theorem to set-up, but do not evaluate, a line integral computing the area of $R$ . $\iint _R\d A = \int _{\answer {\pi }}^{\answer {2\pi }} \answer {-\cos (t)\sin (2t)} \d t$
Set $\vec {H}(x,y) = \vector {-y/2,x/2}$ and use Green’s theorem to set-up, but do not evaluate, a line integral computing the area of $R$ . $\iint _R\d A = \int _{\answer {\pi }}^{\answer {2\pi }} \answer {\cos (2t)\sin (t) - \cos (t)\sin (2t)/2} \d t$

Use one of these integrals to compute the area of one loop of the analemma. $\text {area} = \answer {4/3}$