∫_(−(√3) ) ^(√3) ∫_1 ^(√(4−x^2 )) (x^2 +y^2 )^(3/2) dydx


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Question Number 88710 by jagoll last updated on 12/Apr/20

$\underset{- \sqrt{3}}{\int^{\sqrt{3}}} \underset{1}{\int^{\sqrt{4 - x^{2}}}} {(x^{2} + y^{2})}^{3 / 2} d y d x$
Commented by mathmax by abdo last updated on 12/Apr/20
$I =∫_(−(√3)) ^(√3) (∫_1 ^(√(4−x^2 )) (x^2 +y^2 )^(3/2) dy)dx =∫_(−(√3)) ^(√3) A(x)dx A(x) =∫_1 ^(√(4−x^2 )) (y^2 +x^2 )^(3/2) dy =_(y=∣x∣sh(t)) ∫_(argsh((1/(∣x∣)))) ^(argsh(((√(4−x^2 ))/(∣x∣)))) x^3 (ch^2 t)^(3/2) ∣x∣ch(t)dt =∫_(ln((1/(∣x∣))+(√(1+(1/x^2 ))))) ^(ln(((√(4−x^2 ))/(∣x∣))+(√(1+((4−x^2 )/x^2 ))))) x^3 ∣x∣ ch^4 t dt =x^3 ∣x∣ ∫_(α(x)) ^(β(x)) (((1+ch(2t))/2))^2 dt =(1/4)x^3 ∣x∣∫_(α(x)) ^(β(x)) (1+2ch(2t)+ch^2 (2t))dt =(1/4)x^3 ∣x∣(β(x)−α(x))+(1/4)[sh(2t)]_(α(x)) ^(β(x)) +(1/4)x^3 ∣x∣ ∫_(α(x)) ^(β(x)) (((1+ch(4t))/2))dt =(1/4)x^3 ∣x∣(β(x)−α(x))+(1/4)[((e^(2t) −e^(−2t) )/2)]_(α(x)) ^(β(x)) +(1/8)x^3 ∣x∣(β(x)−α(x)) +(1/(32))x^3 ∣x∣ [sh(4t)]_(α(x)) ^(β(x)) =(3/8)x^3 ∣x∣(β(x)−α(x))+(1/8){e^(2β(x)) −e^(−2α(x)) −e^(2α(x)) +e^(−2α(x)) } +(1/(64))x^3 ∣x∣{e^(4β(x)) −e^(−4β(x)) −e^(4α(x)) +e^(−4α(x)) }....be continued...$
$I = \int_{- \sqrt{3}}^{\sqrt{3}} (\int_{1}^{\sqrt{4 - x^{2}}} {(x^{2} + y^{2})}^{\frac{3}{2}} d y) d x = \int_{- \sqrt{3}}^{\sqrt{3}} A (x) d x$ $A (x) = \int_{1}^{\sqrt{4 - x^{2}}} {(y^{2} + x^{2})}^{\frac{3}{2}} d y =_{y =∣ x ∣ s h (t)} \int_{a r g s h (\frac{1}{∣ x ∣})}^{a r g s h (\frac{\sqrt{4 - x^{2}}}{∣ x ∣})} x^{3} {({c h}^{2} t)}^{\frac{3}{2}} ∣ x ∣ c h (t) d t$ $= \int_{l n (\frac{1}{∣ x ∣} + \sqrt{1 + \frac{1}{x^{2}}})}^{l n (\frac{\sqrt{4 - x^{2}}}{∣ x ∣} + \sqrt{1 + \frac{4 - x^{2}}{x^{2}}})} x^{3} ∣ x ∣ {c h}^{4} t d t$ $= x^{3} ∣ x ∣ \int_{α (x)}^{β (x)} {(\frac{1 + c h (2 t)}{2})}^{2} d t = \frac{1}{4} x^{3} ∣ x ∣ \int_{α (x)}^{β (x)} (1 + 2 c h (2 t) + {c h}^{2} (2 t)) d t$ $= \frac{1}{4} x^{3} ∣ x ∣ (β (x) - α (x)) + \frac{1}{4} {[s h (2 t)]}_{α (x)}^{β (x)} + \frac{1}{4} x^{3} ∣ x ∣ \int_{α (x)}^{β (x)} (\frac{1 + c h (4 t)}{2}) d t$ $= \frac{1}{4} x^{3} ∣ x ∣ (β (x) - α (x)) + \frac{1}{4} {[\frac{e^{2 t} - e^{- 2 t}}{2}]}_{α (x)}^{β (x)} + \frac{1}{8} x^{3} ∣ x ∣ (β (x) - α (x))$ $+ \frac{1}{32} x^{3} ∣ x ∣ {[s h (4 t)]}_{α (x)}^{β (x)}$ $= \frac{3}{8} x^{3} ∣ x ∣ (β (x) - α (x)) + \frac{1}{8} {e^{2 β (x)} - e^{- 2 α (x)} - e^{2 α (x)} + e^{- 2 α (x)}}$ $+ \frac{1}{64} x^{3} ∣ x ∣ {e^{4 β (x)} - e^{- 4 β (x)} - e^{4 α (x)} + e^{- 4 α (x)}} . . . . b e c o n t i n u e d . . .$
	Answered by john santu last updated on 12/Apr/20
	$let me try. to converting to polar coordinates y =(√(4−x^2 )) ⇒ { ((y=1)),((r=2 , θ = (π/6), ((5π)/6))) :} I = ∫_(π/6) ^((5π)/6) ∫_(csc θ) ^2 (r^2 )^(3/2) (r dr dθ) I = ∫_(π/6) ^((5π)/6) ∫_(csc θ) ^2 r^4 dr dθ I = ∫_(π/6) ^((5π)/6) {(1/5)r^5 ]_(csc θ) ^2 } dθ I = (1/5)∫_(π/6) ^((5π)/6) (32−csc^5 θ ) dθ I = (1/5)(((64π)/3) −∫_(π/6) ^((5π)/6) csc^5 θ dθ) I = ((64π)/(15)) −(1/5) [−(1/4) csc^3 θ cot θ − (3/8)csc θ cot θ −(3/8) ln ∣ csc θ + cot θ ∣ ]^((5π)/6) _( (π/6)) I = (1/(60)) ( 256π −9 ln (2+(√3) )−66(√3) )$
	$l e t m e t r y .$ $t o c o n v e r t i n g t o p o l a r c o o r d i n a t e s$ $y = \sqrt{4 - x^{2}} \Rightarrow {\begin{cases} y = 1 \\ r = 2, θ = \frac{π}{6}, \frac{5 π}{6} \end{cases}$ $I = \underset{\frac{π}{6}}{\int^{\frac{5 π}{6}}} \underset{c s c θ}{\int^{2}} {(r^{2})}^{\frac{3}{2}} (r d r d θ)$ $I = \underset{\frac{π}{6}}{\int^{\frac{5 π}{6}}} \underset{c s c θ}{\int^{2}} r^{4} d r d θ$ $I = \underset{\frac{π}{6}}{\int^{\frac{5 π}{6}}} {{\frac{1}{5} r^{5}]}_{c s c θ}^{2}} d θ$ $I = \frac{1}{5} \underset{\frac{π}{6}}{\int^{\frac{5 π}{6}}} (32 - c s c^{5} θ) d θ$ $I = \frac{1}{5} (\frac{64 π}{3} - \underset{\frac{π}{6}}{\int^{\frac{5 π}{6}}} c s c^{5} θ d θ)$ $I = \frac{64 π}{15} - \frac{1}{5} [- \frac{1}{4} c s c^{3} θ \cot θ$ $Missing \left or extra \right$ $I = \frac{1}{60} (256 π - 9 \ln (2 + \sqrt{3}) - 66 \sqrt{3})$
	Commented by jagoll last updated on 12/Apr/20

	$t h a n k y o u s i r .$
	Answered by mr W last updated on 12/Apr/20

	$I = \underset{- \sqrt{3}}{\int^{\sqrt{3}}} \underset{1}{\int^{\sqrt{4 - x^{2}}}} {(x^{2} + y^{2})}^{3 / 2} d y d x$ $= 2 \underset{0}{\int^{\sqrt{3}}} \underset{1}{\int^{\sqrt{4 - x^{2}}}} {(x^{2} + y^{2})}^{3 / 2} d y d x$ $= 2 \int_{\frac{π}{6}}^{\frac{π}{2}} \int_{\frac{1}{\sin θ}}^{2} r^{4} d r d θ$ $= \frac{2}{5} \int_{\frac{π}{6}}^{\frac{π}{2}} (2^{5} - \frac{1}{\sin^{5} θ}) d θ$ $= \frac{2}{5} [32 (\frac{π}{2} - \frac{π}{6}) - \int_{\frac{π}{6}}^{\frac{π}{2}} \frac{d θ}{\sin^{5} θ}]$ $= \frac{2}{5} [\frac{32 π}{3} - \int_{\frac{π}{6}}^{\frac{π}{2}} \frac{d θ}{\sin^{5} θ}]$ $= \frac{2}{5} [\frac{32 π}{3} - \frac{11 \sqrt{3}}{4} - \frac{3 \ln (2 + \sqrt{3})}{8}]$ $= \frac{1}{60} [256 π - 66 \sqrt{3} - 9 \ln (2 + \sqrt{3})]$ $\approx 11.301$ $\int_{\frac{π}{6}}^{\frac{π}{2}} \frac{d θ}{\sin^{5} θ}$ $= \int_{\frac{π}{6}}^{\frac{π}{2}} \frac{\sin θ d θ}{\sin^{6} θ}$ $= - \int_{\frac{π}{6}}^{\frac{π}{2}} \frac{d (\cos θ)}{{(1 - \cos^{2} θ)}^{3}}$ $= \int_{0}^{\frac{\sqrt{3}}{2}} \frac{d u}{{(1 - u^{2})}^{3}}$ $. . . . . .$ $= \frac{3}{8} \ln (2 + \sqrt{3}) + \frac{11 \sqrt{3}}{4}$
	Commented by mr W last updated on 12/Apr/20

	${\begin{cases} θ = \frac{π}{6} \\ θ = \frac{5 π}{6} \end{cases}$ $e q u a l t o 2 t i m e s θ f r o m \frac{π}{6} t o \frac{π}{2} d u e t o$ $s y m m e t r y$
	Commented by john santu last updated on 12/Apr/20

	$s i r, i f r = 2, y = 1$ $t h e n 2 \sin θ = 1 {\begin{cases} θ = \frac{π}{6} \\ θ = \frac{5 π}{6} \end{cases}$
	Commented by mr W last updated on 12/Apr/20

	$o u r r e s u l t s a r e t h e s a m e .$
	Commented by john santu last updated on 12/Apr/20

	$o o y e s s i r . t h a n k y o u$
	Commented by jagoll last updated on 12/Apr/20

	$t h a n k y o u m r W$
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