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:

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