19)
x = 4 + y 2 x = 2 y 2 {\displaystyle {\begin{aligned}&\color {red}\mathbf {x=4+y^{2}} &\color {royalblue}\mathbf {x=2y^{2}} \\\end{aligned}}}
4 + y 2 = 2 y 2 4 = y 2 y = 4 y = ± 2 ∫ − 2 2 [ ( 4 + y 2 ) − ( 2 y 2 ) ] d y = ∫ − 2 2 [ 4 − y 2 ] d y = 4 y − y 3 3 | − 2 2 = 4 ( 2 ) − ( 2 ) 3 3 − ( 4 ( − 2 ) − ( − 2 ) 3 3 ) = 8 − 8 3 − ( − 8 + 8 3 ) = 16 − 16 3 = 48 3 − 16 3 = 32 3 {\displaystyle {\begin{aligned}4+y^{2}&=2y^{2}\\4&=y^{2}\\y&={\sqrt {4}}\\y&=\pm 2\\\int _{-2}^{2}[(4+y^{2})-(2y^{2})]dy\\=\int _{-2}^{2}[4-y^{2}]dy\\=4y-{\frac {y^{3}}{3}}|_{-2}^{2}\\=4(2)-{\frac {(2)^{3}}{3}}-(4(-2)-{\frac {(-2)^{3}}{3}})\\=8-{\frac {8}{3}}-(-8+{\frac {8}{3}})\\=16-{\frac {16}{3}}\\={\frac {48}{3}}-{\frac {16}{3}}\\={\frac {32}{3}}\\\end{aligned}}}
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![{\displaystyle {\begin{aligned}4+y^{2}&=2y^{2}\\4&=y^{2}\\y&={\sqrt {4}}\\y&=\pm 2\\\int _{-2}^{2}[(4+y^{2})-(2y^{2})]dy\\=\int _{-2}^{2}[4-y^{2}]dy\\=4y-{\frac {y^{3}}{3}}|_{-2}^{2}\\=4(2)-{\frac {(2)^{3}}{3}}-(4(-2)-{\frac {(-2)^{3}}{3}})\\=8-{\frac {8}{3}}-(-8+{\frac {8}{3}})\\=16-{\frac {16}{3}}\\={\frac {48}{3}}-{\frac {16}{3}}\\={\frac {32}{3}}\\\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/576a4e8297c129d74aefe79ca36d6a7fc901ec05)
