Question

Изменить порядок интегрирования и вычислить двойной интеграл

$intlimits^2_{-2} {} , dx intlimits^{frac{sqrt{4-x^{2} } }{sqrt{2} } }_{-frac{sqrt{4-x^{2} } }{sqrt{2} } } {} , dy$

Answer 1

Ответ: 2π√2

Объяснение:

$-2leqslant xleqslant 2,\ -dfrac{sqrt{4-x^2}}{sqrt{2}}leqslant yleqslantdfrac{sqrt{4-x^2}}{sqrt{2}}$

$dfrac{sqrt{4-x^2}}{sqrt{2}}=y~~~Longleftrightarrow~~~~ 4-x^2=2y^2~~~~Longleftrightarrow~~~~ x=pmsqrt{4-2y^2}$

Найдем точки пересечения с осью ординат

x = 0:  4 - 2y² = 0    ⇔     y = ± √2

$displaystyle intlimits^2_{-2} {} , dxintlimits^{dfrac{sqrt{4-x^2}}{sqrt{2}}}_{-dfrac{sqrt{4-x^2}}{sqrt{2}}} {} , dy=intlimits^{sqrt{2}}_{-sqrt{2}} {} , dyintlimits^big{sqrt{4-2y^2}}_big{-sqrt{4-2y^2}} {} , dx=\ \ \ \ =intlimits^{sqrt{2}}_{-sqrt{2}} dy,,, xbigg|^{sqrt{4-2y^2}}_{-sqrt{4-2y^2}}=intlimits^{sqrt{2}}_{-sqrt{2}}bigg(sqrt{4-2y^2}+sqrt{4-2y^2}bigg)dy=$

$=displaystyle 2intlimits^{sqrt{2}}_{-sqrt{2}}sqrt{4-2y^2}dy=2cdotdfrac{1}{sqrt{2}}bigg(ysqrt{2-y^2}+2arcsinfrac{y}{sqrt{2}}bigg)bigg|^{sqrt{2}}_{-sqrt{2}}=\ \ \ =2sqrt{2}bigg(arcsin1-arcsin(-1)bigg)=2sqrt{2}bigg[dfrac{pi}{2}-bigg(-dfrac{pi}{2}bigg)bigg]=2pisqrt{2}$