e96de4…: “Under a proud bird”

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Consider an astroid:

This curve is drawn in a counterclockwise fashion by $\vec {a}(t) = \vector {\cos ^3(t), \sin ^3(t)}$ as $t$ runs from $0$ 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 $R$ . $\iint _R\d A = \int _{\answer {0}}^{\answer {2\pi }} \answer {3\cos ^4(t) \sin ^2(t)} \d t$
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 $R$ . $\iint _R\d A = \int _{\answer {0}}^{\answer {2\pi }} \answer {3\sin ^4(t) \cos ^2(t)} \d t$
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 $R$ . $\iint _R\d A = \int _{\answer {0}}^{\answer {2\pi }} \answer {(3/2)\cos ^4(t)\sin ^2(t) + (3/2) \cos ^2(t)\sin ^4(t)} \d t$

Use one of these integrals to compute the area of the astroid. $\text {area} = \answer {3\pi /8}$