Question

Calculate the length of the asteroid of x^(2/3)+y^(2/3)=1

Answer 1

Hii

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question as follows. Show that for any planar region ΩΩ,

area(Ω)=12∮∂Ω(xdy−ydx).area(Ω)=12∮∂Ω(xdy−ydx).

Use this result to find the area enclosed by the astroid x2/3+y2/3=a2/3x2/3+y2/3=a2/3 for a>0a>0.

The first part's easy:

RHS======12∮∂Ω(xdy−ydx)12∮∂Ω(−ydx+xdy)12∬Ω(1+1)dA∬Ω1dAarea(Ω)LHS.RHS=12∮∂Ω(xdy−ydx)=12∮∂Ω(−ydx+xdy)=12∬Ω(1+1)dA=∬Ω1dA=area(Ω)=LHS.

But the next part really has me stumped. In this case Ω={(x,y):x2/3+y2/3≤a2/3}Ω={(x,y):x2/3+y2/3≤a2/3} and

area(Ω)=∬Ω1dA=12∮∂Ω(xdy−ydx)
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