Question Number 34312 by prof Abdo imad last updated on 03/May/18
$${calculate}\:{I}\:\:=\:\int\int_{{D}} {x}^{\mathrm{3}} {dxdy}\:\:\:{on}\:{the}\:{domain} \\ $$$${D}\:=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /\mathrm{1}\leqslant{x}\leqslant\mathrm{2}\:,\:{x}^{\mathrm{2}} −{y}^{\mathrm{2}} −\mathrm{1}\geqslant\mathrm{0}\right\} \\ $$
Commented by math khazana by abdo last updated on 05/May/18
![x^2 −y^2 −1≥0 ⇒ x^2 −1 ≥y^2 ⇒y^2 ≤ x^2 −1 ⇒ −(√(x^2 −1)) ≤y≤(√(x^2 −1)) I = ∫_1 ^2 ( ∫_(−(√(x^2 −1))) ^(√(x^2 −1)) dy)x^3 dx = 2 ∫_1 ^2 x^3 (√(x^2 −1))dx changement x=ch(t) ⇔ t argchx =ln( x +(√(x^2 −1))) I = 2 ∫_0 ^(ln(2+(√3))) ch^3 t sh(t)sh(t)dt by parts u=sht and v^′ =sht ch^3 t I =2( [(1/4)sht ch^4 t]_0 ^(ln(2+(√3))) −∫_0 ^(ln(2+(√3))) cht (1/4)ch^4 dt) I =(1/2)( sh(ln(2+(√3)))ch^4 (ln(2+(√3)) −(1/2) ∫_0 ^(ln(2+(√3))) ch^5 t dt but ch^5 t = (((e^t +e^(−t) )/2))^5 =(1/(32)) Σ_(k=0) ^5 C_5 ^k e^(kt) e^((5−k)t) ...](Q34417.png)
$${x}^{\mathrm{2}} \:−{y}^{\mathrm{2}} −\mathrm{1}\geqslant\mathrm{0}\:\Rightarrow\:{x}^{\mathrm{2}} \:−\mathrm{1}\:\geqslant{y}^{\mathrm{2}} \:\Rightarrow{y}^{\mathrm{2}} \:\leqslant\:{x}^{\mathrm{2}} −\mathrm{1} \\ $$$$\Rightarrow\:−\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\leqslant{y}\leqslant\sqrt{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$${I}\:=\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\left(\:\int_{−\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}} ^{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}} \:{dy}\right){x}^{\mathrm{3}} \:{dx} \\ $$$$=\:\mathrm{2}\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:{x}^{\mathrm{3}} \sqrt{{x}^{\mathrm{2}} −\mathrm{1}}{dx}\:\:\:{changement}\:{x}={ch}\left({t}\right)\:\Leftrightarrow \\ $$$${t}\:{argchx}\:={ln}\left(\:{x}\:+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right) \\ $$$${I}\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)} {ch}^{\mathrm{3}} {t}\:\:{sh}\left({t}\right){sh}\left({t}\right){dt}\:{by}\:{parts} \\ $$$${u}={sht}\:{and}\:{v}^{'} \:={sht}\:{ch}^{\mathrm{3}} {t} \\ $$$${I}\:=\mathrm{2}\left(\:\left[\frac{\mathrm{1}}{\mathrm{4}}{sht}\:{ch}^{\mathrm{4}} {t}\right]_{\mathrm{0}} ^{{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)} \:−\int_{\mathrm{0}} ^{{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)} \:{cht}\:\frac{\mathrm{1}}{\mathrm{4}}{ch}^{\mathrm{4}} \:{dt}\right) \\ $$$${I}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\:{sh}\left({ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\right){ch}^{\mathrm{4}} \left({ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)} {ch}^{\mathrm{5}} {t}\:{dt}\right.\right. \\ $$$${but}\:\:{ch}^{\mathrm{5}} {t}\:=\:\left(\frac{{e}^{{t}} \:+{e}^{−{t}} }{\mathrm{2}}\right)^{\mathrm{5}} \:=\frac{\mathrm{1}}{\mathrm{32}}\:\sum_{{k}=\mathrm{0}} ^{\mathrm{5}} \:{C}_{\mathrm{5}} ^{{k}} \:\:{e}^{{kt}} \:{e}^{\left(\mathrm{5}−{k}\right){t}} \:... \\ $$