find the area of the region bound by the graph of y^2=x^2-x^4

Question

what would the bounds be?

Answer 1

y²=x²-x⁴ is even (evaluates the same for x and -x) and y=±x√(1-x²), so it has symmetry about both axes.

Also, 1-x²≥0 for a real square root, so 1≥x²⇒-1≤x≤1, so the bounds for x are [-1,1]. The x intercepts are (-1,0), (0,0) and (1,0). The figure has a “bow-tie” appearance.

Because of double symmetry we only need to find the definite integral for a quarter of the area:

∫¹₀ydx=∫¹₀x√(1-x²)dx.

Let u=1-x², du=-2xdx, so xdx=-du/2 and the limits for u remain as for x, but interchanged.

-½∫⁰₁√udu=-½(⅔√u³)|⁰₁=⅓. We multiply by 4, so the total area of the figure is ⁴⁄₃ square units.