tex]...

March 2022 1 14 Report

помогите пожалуйста очень срочно!
∫∫[tex] \sqrt{\frac{1-x^2-y^2 }{1+x^2+y^2} } [/tex]dxdy
x[tex] x^{2} + y^{2} \leq 1[/tex]
[tex]x \geq 0[/tex]
[tex]y \geq 0[/tex]

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NNNLLL54

Verified answer

$\iint _{D} \, \sqrt{\frac{1-x^2-y^2}{1+x^2+y^2} }dx\, dy=[\, x=r\cdot cos\varphi ,\; y=rsin\varphi ,\; dx\, dy=r\, dr\, d\varphi \, ]=\\\\=\int _0^{\frac{\pi}{2}}d\varphi \int _0^1\sqrt{\frac{1-r^2}{1+r^2}} \cdot r\, dr=I\\\\\int\sqrt{\frac{1-r^2}{1+r^2}}\cdot r\, dr=[t=r^2,dt=2rdr]=$

$=\int\sqrt{\frac{1-t}{1+t}}\cdot \frac{1}{2}dt=$

$=\frac{1}{2} \int \frac{\sqrt{(1-t)(1-t)}}{\sqrt{(1+t)(1-t)}}dt=\frac{1}{2}\int \frac{1-t}{\sqrt{1-t^2}}dt=\frac{1}{2}\int \frac{dt}{\sqrt{1-t^2}}-\frac{1}{4}\int \frac{2tdt}{\sqrt{1-t^2}}=\\\\=\frac{1}{2}arcsint+\frac{1}{4}*2\sqrt{1-t^2}+C=\frac{1}{2}arcsin x^2+\frac{1}{2}\sqrt{1-x^4}+C$

$I=\int _0^{\frac{\pi}{2}}(\frac{1}{2}arcsinx^2|_0^1+\frac{1}{2}\sqrt{1-x^4}|_0^1)d\varphi =\\\\=\int _0^{\frac{\pi}{2}}(\frac{1}{2}*\frac{\pi}{2}-\frac{1}{2})d\varphi=(\frac{\pi}{4}-\frac{1}{2})\varphi |_0^1=\frac{\pi}{4}-\frac{1}{2}\\\\==$

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Answers & Comments

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sostavte molekulyarnye i ionnye uravneniya reakcij gde eto vozmozhno s pomoshyu

pomogite reshit 27vo2stepeni umnozhenie na 9 v 4 stepeni i razdelit na 81 vo 2st

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