∫ ∫ (√(x^2 + y^2 )) dx dy, (√(3y)) ≤ x ≤ (√(4 − y^2 )) , 0 ≤ y ≤ 2


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Question Number 48725 by Tawa1 last updated on 27/Nov/18

$\int \int \sqrt{x^{2} + y^{2}} dx dy, \sqrt{3 y} ⩽ x ⩽ \sqrt{4 - y^{2}}, 0 ⩽ y ⩽ 2$
Commented by Abdo msup. last updated on 27/Nov/18
$let I =∫∫_D (√(x^2 +y^2 ))dxdy with D={(x,y)∈R^2 /0≤y≤2 and (√(3y))≤x≤(√(4−y^2 )) I =∫_0 ^2 (∫_(√(3y)) ^(√(4−y^2 )) (√(x^2 +y^2 ))dx)dy let find A(y)=∫_(√(3y)) ^(√(4−y^2 )) (√(x^2 +y^2 ))dx A(y) =_(x=ysh(t)) ∫_(argsh(((√3)/(√y)))) ^(argsh(((√(4−y^2 ))/y))) ych(t)y ch(t)dt = y^2 ∫_(ln(((√3)/(√y)) +(√(3/y^2 ))+1)) ^(ln(((√(4−y^2 ))/y) +(√((4−y^2 )/y^2 ))+1)) ((1+ch(2t))/2) dt =(y^2 /2){ln(((√(4−y^2 ))/y)+(2/y))−ln(((√3)/(√y)) +(√(3/y^2 ))+1)} +(y^2 /4)[sh(2t)]_(α(y)) ^(β(y)) =...+(y^2 /8)[ e^(2t) −e^(−2t) ]_(α(y)) ^(β(y)) =....+(y^2 /8){ (((√(4−y^2 ))/y) +(2/y))^2 −(((√3)/(√y)) +(√((3/y^2 )+1)))^(−2) } ⇒ A(y)=(y^2 /2){ln(((√(4−y^2 ))/y) +(2/y))−ln(((√3)/(√y)) +(√((3/y^2 )+1))) +(y^2 /8){ln(((√(4−y^2 ))/y) +(2/y))−ln(((√3)/(√y)) +(√((3/y^2 )+1)))^(−2) } ⇒ I =∫_0 ^2 A(y)dy ...be continued...$
$l e t I = \int \int_{D} \sqrt{x^{2} + y^{2}} d x d y w i t h D = {(x, y) \in R^{2} / 0 ⩽ y ⩽ 2 a n d$ $\sqrt{3 y} ⩽ x ⩽ \sqrt{4 - y^{2}}$ $I = \int_{0}^{2} (\int_{\sqrt{3 y}}^{\sqrt{4 - y^{2}}} \sqrt{x^{2} + y^{2}} d x) d y l e t f i n d A (y) = \int_{\sqrt{3 y}}^{\sqrt{4 - y^{2}}} \sqrt{x^{2} + y^{2}} d x$ $A (y) =_{x = y s h (t)} \int_{a r g s h (\frac{\sqrt{3}}{\sqrt{y}})}^{a r g s h (\frac{\sqrt{4 - y^{2}}}{y})} y c h (t) y c h (t) d t$ $= y^{2} \int_{l n (\frac{\sqrt{3}}{\sqrt{y}} + \sqrt{\frac{3}{y^{2}}} + 1)}^{l n (\frac{\sqrt{4 - y^{2}}}{y} + \sqrt{\frac{4 - y^{2}}{y^{2}}} + 1)} \frac{1 + c h (2 t)}{2} d t$ $= \frac{y^{2}}{2} {l n (\frac{\sqrt{4 - y^{2}}}{y} + \frac{2}{y}) - l n (\frac{\sqrt{3}}{\sqrt{y}} + \sqrt{\frac{3}{y^{2}}} + 1)}$ $+ \frac{y^{2}}{4} {[s h (2 t)]}_{α (y)}^{β (y)} = . . . + \frac{y^{2}}{8} {[e^{2 t} - e^{- 2 t}]}_{α (y)}^{β (y)}$ $= . . . . + \frac{y^{2}}{8} {{(\frac{\sqrt{4 - y^{2}}}{y} + \frac{2}{y})}^{2} - {(\frac{\sqrt{3}}{\sqrt{y}} + \sqrt{\frac{3}{y^{2}} + 1})}^{- 2}} \Rightarrow$ $A (y) = \frac{y^{2}}{2} {l n (\frac{\sqrt{4 - y^{2}}}{y} + \frac{2}{y}) - l n (\frac{\sqrt{3}}{\sqrt{y}} + \sqrt{\frac{3}{y^{2}} + 1})$ $+ \frac{y^{2}}{8} {l n (\frac{\sqrt{4 - y^{2}}}{y} + \frac{2}{y}) - l n {(\frac{\sqrt{3}}{\sqrt{y}} + \sqrt{\frac{3}{y^{2}} + 1})}^{- 2}} \Rightarrow$ $I = \int_{0}^{2} A (y) d y . . . b e c o n t i n u e d . . .$
Commented by Tawa1 last updated on 29/Nov/18

$God bless you sir$
Commented by maxmathsup by imad last updated on 01/Dec/18

$t h a n k y o u s i r .$
	Answered by MJS last updated on 28/Nov/18

	$(1) \int \sqrt{x^{2} + y^{2}} d x =$ $[t = arcsinh \frac{x}{y} \to x = y \sinh t \land d x = y \cosh t d t]$ $= \int y \cosh t \sqrt{y^{2} + y^{2} \sinh^{2} t} d t = y^{2} \int \cosh^{2} t d t =$ $= \frac{y^{2}}{2} \int \cosh 2 t d t + \frac{y^{2}}{2} \int d t = \frac{y^{2}}{4} \sinh 2 t + \frac{y^{2}}{2} t =$ $[y ⩾ 0]$ $= \frac{x}{2} \sqrt{x^{2} + y^{2}} + \frac{y^{2}}{2} arcsinh \frac{x}{y}$ $\underset{\sqrt{3 y}}{\int^{\sqrt{4 - y^{2}}}} \sqrt{x^{2} + y^{2}} d x = \sqrt{4 - y^{2}} + \frac{y^{2}}{2} arcsinh \frac{\sqrt{4 - y^{2}}}{y} - \frac{y}{2} \sqrt{3 (y + 3)} - \frac{y^{2}}{2} arcsinh \sqrt{\frac{3}{y}}$ $(2.1) \int \sqrt{4 - y^{2}} d y =$ $[u = \arcsin \frac{y}{2} \to y = 2 \sin u \land d y = 2 \cos u d u]$ $= \int 2 \cos u \sqrt{4 - {4 \sin}^{2} u} d u = 4 \int \cos^{2} u d u =$ $= 2 \int \cos 2 u d u + 2 \int d u = \sin 2 u + 2 u =$ $= \frac{y}{2} \sqrt{4 - y^{2}} + 2 \arcsin \frac{y}{2}$ $(2.2) \int \frac{y^{2}}{2} arcsinh \frac{\sqrt{4 - y^{2}}}{y} d y =$ $[\begin{matrix} \int f^{'} g = f g - \int {f g}^{'} with : f^{'} = \frac{y^{2}}{2} \to f = \frac{y^{3}}{6} \\ g = arcsinh \frac{\sqrt{4 - y^{2}}}{y} \to g^{'} = - \frac{2}{y \sqrt{4 - y^{2}}} \end{matrix}]$ $= \frac{y^{3}}{6} arcsinh \frac{\sqrt{4 - y^{2}}}{y} + \frac{1}{3} \int \frac{y^{2}}{\sqrt{4 - y^{2}}} d y =$ $\frac{1}{3} \int \frac{y^{2}}{\sqrt{4 - y^{2}}} d y = \frac{1}{3} \int \frac{y^{2} - 4 + 4}{\sqrt{4 - y^{2}}} d y =$ $= - \frac{1}{3} \int \sqrt{4 - y^{2}} d y + \frac{4}{3} \int \frac{d y}{\sqrt{4 - y^{2}}} =$ $= - \frac{y}{6} \sqrt{4 - y^{2}} + \frac{2}{3} \arcsin \frac{y}{2}$ $= \frac{y^{3}}{6} arcsinh \frac{\sqrt{4 - y^{2}}}{y} - \frac{y}{6} \sqrt{4 - y^{2}} + \frac{2}{3} \arcsin \frac{y}{2}$ $(2.3) \int - \frac{y}{2} \sqrt{3 (y + 3)} d y = - \frac{\sqrt{3}}{2} \int y \sqrt{y + 3} d y =$ $[v = y + 3 \to d y = d v]$ $= - \frac{\sqrt{3}}{2} \int (v - 3) \sqrt{v} d v = \frac{3 \sqrt{3}}{2} \int v^{\frac{1}{2}} d v - \frac{\sqrt{3}}{2} \int v^{\frac{3}{2}} d v =$ $= \sqrt{3} v^{\frac{3}{2}} - \frac{\sqrt{3}}{5} v^{\frac{5}{2}} = \sqrt{3} {(y + 3)}^{\frac{3}{2}} - \frac{\sqrt{3}}{5} {(y + 3)}^{\frac{5}{2}} = \frac{\sqrt{3}}{5} (2 - y) {(y + 3)}^{\frac{3}{2}}$ $(2.4) \int - \frac{y^{2}}{2} arcsinh \sqrt{\frac{3}{y}} d y = - \frac{1}{2} \int y^{2} arcsinh \sqrt{\frac{3}{y}} d y =$ $[w = \frac{y^{3}}{3} \to d x = \frac{d w}{y^{2}}]$ $= - \frac{1}{2} \int arcsinh \frac{\sqrt[3]{3}}{\sqrt[6]{w}} d w =$ $[\begin{matrix} \int f^{'} g = f g - \int {f g}^{'} with : f^{'} = 1 \to f = w \\ g = arcsinh \frac{\sqrt[3]{3}}{\sqrt[6]{w}} \to g^{'} = - \frac{\sqrt[3]{3}}{6 w \sqrt{\sqrt[3]{w} + \sqrt[3]{9}}} \end{matrix}]$ $= - \frac{w}{2} arcsinh \frac{\sqrt[3]{3}}{\sqrt[6]{w}} - \frac{\sqrt[3]{3}}{12} \int \frac{d w}{\sqrt{\sqrt[3]{w} + \sqrt[3]{9}}}$ $- \frac{\sqrt[3]{3}}{12} \int \frac{d w}{\sqrt{\sqrt[3]{w} + \sqrt[3]{9}}} =$ $[z = \sqrt[3]{w} + \sqrt[3]{9} \to d w = 3 (z - \sqrt[3]{9}) d z]$ $= - \frac{\sqrt[3]{3}}{4} \int z^{\frac{3}{2}} d z + \frac{3}{2} \int z^{\frac{1}{2}} d z - \frac{3 \sqrt[3]{9}}{4} \int z^{- \frac{1}{2}} d z =$ $= - \frac{\sqrt[3]{3}}{10} z^{\frac{5}{2}} + z^{\frac{3}{2}} - \frac{3 \sqrt[3]{9}}{2} z^{\frac{1}{2}} = (- \frac{\sqrt[3]{3}}{10} \sqrt[3]{w^{2}} + \frac{2}{5} \sqrt[3]{w} - \frac{4 \sqrt[3]{9}}{5}) \sqrt{\sqrt[3]{w} + \sqrt[3]{9}}$ $= - \frac{w}{2} arcsinh \frac{\sqrt[3]{3}}{\sqrt[6]{w}} - (\frac{\sqrt[3]{3}}{10} \sqrt[3]{w^{2}} - \frac{2}{5} \sqrt[3]{w} + \frac{4 \sqrt[3]{9}}{5}) \sqrt{\sqrt[3]{w} + \sqrt[3]{9}} =$ $= - \frac{y^{3}}{6} arcsinh \frac{\sqrt[3]{9}}{y} - \frac{1}{30} (y^{2} - 4 y + 24) \sqrt{3 (y + 3)}$ $\underset{0}{\int^{2}} (\underset{\sqrt{3 y}}{\int^{\sqrt{4 - y^{2}}}} \sqrt{x^{2} + y^{2}} d x) d y = \frac{4 π}{3} - \frac{4}{3} arcsinh \frac{\sqrt[3]{9}}{2} - \frac{2 \sqrt{15}}{3} - \frac{6}{5} \approx - . 805740$
	Commented by MJS last updated on 28/Nov/18

	$. . . please someone check for typos . {it}^{'} s 3 a . m .$ $and I^{'} m really tired . . .$
	Commented by maxmathsup by imad last updated on 28/Nov/18

	$t h a n k y o u s i r f o r t h i s h a r d w o r k t h i s i s t h e w a y .$
	Commented by MJS last updated on 28/Nov/18

	${you}^{'} re welcome$
	Commented by Tawa1 last updated on 29/Nov/18

	$God bless you sir$
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