The area bounded by the curve $$y^2 = x^2 - x^4$$ is

Since we see that both y and x has even terms, it implies the curve is symmetric along x-axis as well as y-axis.

Required area = 4*(area in quadrant I)

Area in quadrant I is given by 

A =$$_0\int^1\sqrt{\ x^2-x^4}dx\ $$

A = $$_0\int^1x\sqrt{1-x^2}dx$$

take $$1-x^2\ =\ t^2$$, then $$-2x\ dx\ =\ 2\ t\ dt$$. when x =0 then t = 1 and  when x = 1 then t = 0

Updated integral can be written as 

$$\int_0^1t^{2\ }dt$$ = $$\left[\frac{t^3\ }{3}\right]_0^1\ =\ \ \frac{\ 1}{3}$$

Required area = 4*A = $$\ \frac{\ 4}{3}$$