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
$calculate I = ∫∫_D x^3 dxdy on the domain D ={(x,y)∈R^2 /1≤x≤2 , x^2 −y^2 −1≥0}$
$c a l c u l a t e I = \int \int_{D} x^{3} d x d y o n t h e d o m a i n$ $D = {(x, y) \in R^{2} / 1 ⩽ x ⩽ 2, x^{2} - y^{2} - 1 ⩾ 0}$
Commented by math khazana by abdo last updated on 05/May/18

$x^{2} - y^{2} - 1 ⩾ 0 \Rightarrow x^{2} - 1 ⩾ y^{2} \Rightarrow y^{2} ⩽ x^{2} - 1$ $\Rightarrow - \sqrt{x^{2} - 1} ⩽ y ⩽ \sqrt{x^{2} - 1}$ $I = \int_{1}^{2} (\int_{- \sqrt{x^{2} - 1}}^{\sqrt{x^{2} - 1}} d y) x^{3} d x$ $= 2 \int_{1}^{2} x^{3} \sqrt{x^{2} - 1} d x c h a n g e m e n t x = c h (t) \Leftrightarrow$ $t a r g c h x = l n (x + \sqrt{x^{2} - 1})$ $I = 2 \int_{0}^{l n (2 + \sqrt{3})} {c h}^{3} t s h (t) s h (t) d t b y p a r t s$ $u = s h t a n d v^{^{'}} = s h t {c h}^{3} t$ $I = 2 ({[\frac{1}{4} s h t {c h}^{4} t]}_{0}^{l n (2 + \sqrt{3})} - \int_{0}^{l n (2 + \sqrt{3})} c h t \frac{1}{4} {c h}^{4} d t)$ $I = \frac{1}{2} (s h (l n (2 + \sqrt{3})) {c h}^{4} (l n (2 + \sqrt{3}) - \frac{1}{2} \int_{0}^{l n (2 + \sqrt{3})} {c h}^{5} t d t$ $b u t {c h}^{5} t = {(\frac{e^{t} + e^{- t}}{2})}^{5} = \frac{1}{32} \sum_{k = 0}^{5} C_{5}^{k} e^{k t} e^{(5 - k) t} . . .$
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