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calculate-I-D-x-3-dxdy-on-the-domain-D-x-y-R-2-1-x-2-x-2-y-2-1-0-

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Question Number 34312 by prof Abdo imad last updated on 03/May/18
calculate I = ∫∫_D x^3 dxdy on the domain D ={(x,y)∈R^2 /1≤x≤2 , x^2 −y^2 −1≥0}
calculateI=∫∫Dx3dxdyonthedomainD={(x,y)∈R2/1⩽x⩽2,x2−y2−1⩾0}
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) ...
x2−y2−1⩾0⇒x2−1⩾y2⇒y2⩽x2−1⇒−x2−1⩽y⩽x2−1I=∫12(∫−x2−1x2−1dy)x3dx=2∫12x3x2−1dxchangementx=ch(t)⇔targchx=ln(x+x2−1)I=2∫0ln(2+3)ch3tsh(t)sh(t)dtbypartsu=shtandv′=shtch3tI=2([14shtch4t]0ln(2+3)−∫0ln(2+3)cht14ch4dt)I=12(sh(ln(2+3))ch4(ln(2+3)−12∫0ln(2+3)ch5tdtbutch5t=(et+e−t2)5=132∑k=05C5kekte(5−k)t…

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