x = 1 − y 2 , x = y 2 − 1 A = ∫ a b [ x R − x L ] d y = ∫ 1 − 1 [ ( 1 − y 2 ) − ( y 2 − 1 ) ] d y = ∫ 1 − 1 [ 2 − 2 y 2 ] d y = [ 2 y − 2 ( y 3 3 ) | − 1 0 = 2 ( 1 ) − 2 ( ( 1 ) 3 3 ) − [ 2 ( − 1 ) − 2 ( ( − 1 ) 3 3 ) ] = 2 − 2 3 + 2 − 2 3 = 4 − 4 3 = 8 3 {\displaystyle {\begin{aligned}x&=1-y^{2},x=y^{2}-1\\[1ex]A&=\int _{a}^{b}[x_{R}-x_{L}]dy=\int _{1}^{-1}[(1-y^{2})-(y^{2}-1)]dy\\[2ex]&=\int _{1}^{-1}[2-2y^{2}]dy=[2y-2({\frac {y^{3}}{3}})|_{-1}^{0}\\[2ex]&=2(1)-2({\frac {(1)^{3}}{3}})-[2(-1)-2({\frac {(-1)^{3}}{3}})]\\[2ex]&=2-{\frac {2}{3}}+2-{\frac {2}{3}}=4-{\frac {4}{3}}\\[2ex]&={\frac {8}{3}}\end{aligned}}}
![{\displaystyle {\begin{aligned}x&=1-y^{2},x=y^{2}-1\\[1ex]A&=\int _{a}^{b}[x_{R}-x_{L}]dy=\int _{1}^{-1}[(1-y^{2})-(y^{2}-1)]dy\\[2ex]&=\int _{1}^{-1}[2-2y^{2}]dy=[2y-2({\frac {y^{3}}{3}})|_{-1}^{0}\\[2ex]&=2(1)-2({\frac {(1)^{3}}{3}})-[2(-1)-2({\frac {(-1)^{3}}{3}})]\\[2ex]&=2-{\frac {2}{3}}+2-{\frac {2}{3}}=4-{\frac {4}{3}}\\[2ex]&={\frac {8}{3}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/78933015cdc8a98d15931cfb0cb7cdb3c0c0ca6e)
