find the area between the curves y=x^2 and y=1-x^2

find the area between the curves y=x^2 and y=1-x^2

I. limits above and below the integral sign:

The Cut of two functions :

$$\\ y_c = x_c^2 = 1-x_c^2 \\ 2x_c^2=1 \\ x_c^2=\frac{1}{2} \\ x_c=\pm\sqrt{\frac{1}{2}}\\ x_c=\pm\frac{\sqrt{2}}{2}$$  

II. The area between is:

  $$\small{\text{ $ \int\limits_{ x=-\frac{\sqrt{2}}{2} }^{ \frac{\sqrt{2}}{2} }(1-x^2)-x^2 \ dx = \int\limits_{x=-\frac{\sqrt{2}}{2} }^{ \frac{\sqrt{2}}{2} } 1-2x^2 \ dx = \int\limits_{x=-\frac{\sqrt{2}}{2} }^{ \frac{\sqrt{2}}{2} } dx - 2\int\limits_{x=-1}^{+1}x^2 \ dx $ }}\\\\ \small{\text{ $ =[x]_{- \frac{\sqrt{2}}{2} }^{ \frac{\sqrt{2}}{2} }-2[\frac{x^3}{3}]_{ -\frac{\sqrt{2}}{2} }^{ \frac{\sqrt{2}}{2} } =[\frac{ \sqrt{2} }{2} -(-\frac{\sqrt{2}}{2} )]-2[\frac{ (\frac{\sqrt{2}}{2})^3}{3}-\frac{(-(\frac{\sqrt{2}}{2}) )^3}{3}] =\sqrt{2} - 2[\frac{4*\sqrt{2}}{24}-\frac{-4*\sqrt{2}}{24}] $ }}\\\\ \small{\text{ $ =\sqrt{2} -2[ \frac{2*4 \sqrt{2} }{24}] =\sqrt{2} - \frac{2}{3} \sqrt{2} =\frac{1}{3} \sqrt{2} $ }} \\\\ \small{\text{The area is $ \frac{1}{3} \sqrt{2} $}}$$