Answer
$A = \frac{4}{3}$
Work Step by Step
$$\eqalign{
& {y^2} = {x^2}\left( {1 - {x^2}} \right) \cr
& {\text{We can calculate the area of the region shown below}}{\text{, then}} \cr
& {\text{the total area of the region is }}4{\text{ times that area}}{\text{.}} \cr
& {\text{Solving the implicit function for }}y \cr
& y = \pm \sqrt {{x^2}\left( {1 - {x^2}} \right)} \cr
& y = \pm x\sqrt {1 - {x^2}} \cr
& {\text{The area is in the quadrant I}}{\text{, then }}y > 0{\text{ and }}x > 0 \cr
& y = x\sqrt {1 - {x^2}} \cr
& {\text{The total area is given by}} \cr
& A = 4\int_0^1 {x\sqrt {1 - {x^2}} } dx \cr
& {\text{Rewrite}} \cr
& A = 2\int_1^0 {\left( { - 2x} \right)\sqrt {1 - {x^2}} } dx \cr
& {\text{Integrating}} \cr
& A = 2\left[ {\frac{{2{{\left( {1 - {x^2}} \right)}^{3/2}}}}{3}} \right]_1^0 \cr
& A = \frac{4}{3}\left[ {{{\left( {1 - {0^2}} \right)}^{3/2}} - {{\left( {1 - {1^2}} \right)}^{3/2}}} \right] \cr
& A = \frac{4}{3}\left[ {{{\left( 1 \right)}^{3/2}} - {{\left( 0 \right)}^{3/2}}} \right] \cr
& A = \frac{4}{3} \cr} $$
