Question Number 88710 by jagoll last updated on 12/Apr/20
$$\underset{−\sqrt{\mathrm{3}}\:} {\overset{\sqrt{\mathrm{3}}} {\int}}\underset{\mathrm{1}} {\overset{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }} {\int}}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} \:{dydx} \\ $$
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...](Q88728.png)
$${I}\:=\int_{−\sqrt{\mathrm{3}}} ^{\sqrt{\mathrm{3}}} \:\:\:\left(\int_{\mathrm{1}} ^{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:{dy}\right){dx}\:=\int_{−\sqrt{\mathrm{3}}} ^{\sqrt{\mathrm{3}}} \:{A}\left({x}\right){dx} \\ $$$${A}\left({x}\right)\:=\int_{\mathrm{1}} ^{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }} \left({y}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:{dy}\:\:=_{{y}=\mid{x}\mid{sh}\left({t}\right)} \:\:\int_{{argsh}\left(\frac{\mathrm{1}}{\mid{x}\mid}\right)} ^{{argsh}\left(\frac{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}{\mid{x}\mid}\right)} {x}^{\mathrm{3}} \left({ch}^{\mathrm{2}} {t}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \mid{x}\mid{ch}\left({t}\right){dt} \\ $$$$=\int_{{ln}\left(\frac{\mathrm{1}}{\mid{x}\mid}+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}\right)} ^{{ln}\left(\frac{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}{\mid{x}\mid}+\sqrt{\mathrm{1}+\frac{\mathrm{4}−{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} }}\right)} {x}^{\mathrm{3}} \mid{x}\mid\:{ch}^{\mathrm{4}} {t}\:{dt} \\ $$$$={x}^{\mathrm{3}} \mid{x}\mid\:\int_{\alpha\left({x}\right)} ^{\beta\left({x}\right)} \:\:\left(\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}\right)^{\mathrm{2}} \:{dt}\:=\frac{\mathrm{1}}{\mathrm{4}}{x}^{\mathrm{3}} \mid{x}\mid\int_{\alpha\left({x}\right)} ^{\beta\left({x}\right)} \:\:\left(\mathrm{1}+\mathrm{2}{ch}\left(\mathrm{2}{t}\right)+{ch}^{\mathrm{2}} \left(\mathrm{2}{t}\right)\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{x}^{\mathrm{3}} \mid{x}\mid\left(\beta\left({x}\right)−\alpha\left({x}\right)\right)+\frac{\mathrm{1}}{\mathrm{4}}\left[{sh}\left(\mathrm{2}{t}\right)\right]_{\alpha\left({x}\right)} ^{\beta\left({x}\right)} \:+\frac{\mathrm{1}}{\mathrm{4}}{x}^{\mathrm{3}} \mid{x}\mid\:\int_{\alpha\left({x}\right)} ^{\beta\left({x}\right)} \left(\frac{\mathrm{1}+{ch}\left(\mathrm{4}{t}\right)}{\mathrm{2}}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{x}^{\mathrm{3}} \mid{x}\mid\left(\beta\left({x}\right)−\alpha\left({x}\right)\right)+\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{{e}^{\mathrm{2}{t}} −{e}^{−\mathrm{2}{t}} }{\mathrm{2}}\right]_{\alpha\left({x}\right)} ^{\beta\left({x}\right)} \:+\frac{\mathrm{1}}{\mathrm{8}}{x}^{\mathrm{3}} \mid{x}\mid\left(\beta\left({x}\right)−\alpha\left({x}\right)\right) \\ $$$$+\frac{\mathrm{1}}{\mathrm{32}}{x}^{\mathrm{3}} \mid{x}\mid\:\left[{sh}\left(\mathrm{4}{t}\right)\right]_{\alpha\left({x}\right)} ^{\beta\left({x}\right)} \\ $$$$=\frac{\mathrm{3}}{\mathrm{8}}{x}^{\mathrm{3}} \mid{x}\mid\left(\beta\left({x}\right)−\alpha\left({x}\right)\right)+\frac{\mathrm{1}}{\mathrm{8}}\left\{{e}^{\mathrm{2}\beta\left({x}\right)} −{e}^{−\mathrm{2}\alpha\left({x}\right)} −{e}^{\mathrm{2}\alpha\left({x}\right)} +{e}^{−\mathrm{2}\alpha\left({x}\right)} \right\} \\ $$$$+\frac{\mathrm{1}}{\mathrm{64}}{x}^{\mathrm{3}} \mid{x}\mid\left\{{e}^{\mathrm{4}\beta\left({x}\right)} −{e}^{−\mathrm{4}\beta\left({x}\right)} −{e}^{\mathrm{4}\alpha\left({x}\right)} +{e}^{−\mathrm{4}\alpha\left({x}\right)} \right\}....{be}\:{continued}... \\ $$
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) )](Q88730.png)
$${let}\:{me}\:{try}. \\ $$$${to}\:{converting}\:{to}\:{polar}\:{coordinates} \\ $$$${y}\:=\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\:\Rightarrow\begin{cases}{{y}=\mathrm{1}}\\{{r}=\mathrm{2}\:,\:\theta\:=\:\frac{\pi}{\mathrm{6}},\:\frac{\mathrm{5}\pi}{\mathrm{6}}}\end{cases} \\ $$$${I}\:=\:\underset{\frac{\pi}{\mathrm{6}}} {\overset{\frac{\mathrm{5}\pi}{\mathrm{6}}} {\int}}\:\underset{{csc}\:\theta} {\overset{\mathrm{2}} {\int}}\:\left({r}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:\left({r}\:{dr}\:{d}\theta\right) \\ $$$${I}\:=\:\underset{\frac{\pi}{\mathrm{6}}} {\overset{\frac{\mathrm{5}\pi}{\mathrm{6}}} {\int}}\:\underset{{csc}\:\theta} {\overset{\mathrm{2}} {\int}}\:{r}^{\mathrm{4}} \:{dr}\:{d}\theta \\ $$$$\left.{I}\:=\:\underset{\frac{\pi}{\mathrm{6}}} {\overset{\frac{\mathrm{5}\pi}{\mathrm{6}}} {\int}}\:\left\{\frac{\mathrm{1}}{\mathrm{5}}{r}^{\mathrm{5}} \:\right]_{{csc}\:\theta} ^{\mathrm{2}} \right\}\:{d}\theta \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{5}}\underset{\frac{\pi}{\mathrm{6}}} {\overset{\frac{\mathrm{5}\pi}{\mathrm{6}}} {\int}}\:\left(\mathrm{32}−{csc}\:^{\mathrm{5}} \:\theta\:\right)\:{d}\theta \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{5}}\left(\frac{\mathrm{64}\pi}{\mathrm{3}}\:−\underset{\frac{\pi}{\mathrm{6}}} {\overset{\frac{\mathrm{5}\pi}{\mathrm{6}}} {\int}}\:{csc}\:^{\mathrm{5}} \:\theta\:{d}\theta\right) \\ $$$${I}\:=\:\frac{\mathrm{64}\pi}{\mathrm{15}}\:−\frac{\mathrm{1}}{\mathrm{5}}\:\left[−\frac{\mathrm{1}}{\mathrm{4}}\:{csc}\:^{\mathrm{3}} \:\theta\:\mathrm{cot}\:\theta\:\right. \\ $$$$\left.−\:\frac{\mathrm{3}}{\mathrm{8}}{csc}\:\theta\:\mathrm{cot}\:\theta\:−\frac{\mathrm{3}}{\mathrm{8}}\:\mathrm{ln}\:\mid\:{csc}\:\theta\:+\:\mathrm{cot}\:\theta\:\mid\:\underset{\:\:\:\frac{\pi}{\mathrm{6}}} {\right]}^{\frac{\mathrm{5}\pi}{\mathrm{6}}} \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{60}}\:\left(\:\mathrm{256}\pi\:−\mathrm{9}\:\mathrm{ln}\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right)−\mathrm{66}\sqrt{\mathrm{3}}\:\right)\: \\ $$
Commented by jagoll last updated on 12/Apr/20
$${thank}\:{you}\:{sir}.\: \\ $$
Answered by mr W last updated on 12/Apr/20
![I=∫_(−(√3) ) ^(√3) ∫_1 ^(√(4−x^2 )) (x^2 +y^2 )^(3/2) dydx =2∫_( 0 ) ^(√3) ∫_1 ^(√(4−x^2 )) (x^2 +y^2 )^(3/2) dydx =2∫_(π/6) ^(π/2) ∫_(1/(sin θ)) ^( 2) r^4 drdθ =(2/5)∫_(π/6) ^(π/2) (2^5 −(1/(sin^5 θ)))dθ =(2/5)[32((π/2)−(π/6))−∫_(π/6) ^(π/2) (dθ/(sin^5 θ))] =(2/5)[((32π)/3)−∫_(π/6) ^(π/2) (dθ/(sin^5 θ))] =(2/5)[((32π)/3)−((11(√3))/4)−((3 ln (2+(√3)))/8)] =(1/(60))[256π−66(√3)−9 ln (2+(√3))] ≈11.301 ∫_(π/6) ^(π/2) (dθ/(sin^5 θ)) =∫_(π/6) ^(π/2) ((sin θdθ)/(sin^6 θ)) =−∫_(π/6) ^(π/2) ((d(cos θ))/((1−cos^2 θ)^3 )) =∫_0 ^((√3)/2) (du/((1−u^2 )^3 )) ...... =(3/8)ln (2+(√3))+((11(√3))/4)](Q88736.png)
$${I}=\underset{−\sqrt{\mathrm{3}}\:} {\overset{\sqrt{\mathrm{3}}} {\int}}\underset{\mathrm{1}} {\overset{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }} {\int}}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} \:{dydx} \\ $$$$=\mathrm{2}\underset{\:\mathrm{0}\:} {\overset{\sqrt{\mathrm{3}}} {\int}}\underset{\mathrm{1}} {\overset{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }} {\int}}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} \:{dydx} \\ $$$$=\mathrm{2}\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{2}}} \int_{\frac{\mathrm{1}}{\mathrm{sin}\:\theta}} ^{\:\mathrm{2}} {r}^{\mathrm{4}} {drd}\theta \\ $$$$=\frac{\mathrm{2}}{\mathrm{5}}\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{2}^{\mathrm{5}} −\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{5}} \:\theta}\right){d}\theta \\ $$$$=\frac{\mathrm{2}}{\mathrm{5}}\left[\mathrm{32}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{6}}\right)−\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\theta}{\mathrm{sin}^{\mathrm{5}} \:\theta}\right] \\ $$$$=\frac{\mathrm{2}}{\mathrm{5}}\left[\frac{\mathrm{32}\pi}{\mathrm{3}}−\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\theta}{\mathrm{sin}^{\mathrm{5}} \:\theta}\right] \\ $$$$=\frac{\mathrm{2}}{\mathrm{5}}\left[\frac{\mathrm{32}\pi}{\mathrm{3}}−\frac{\mathrm{11}\sqrt{\mathrm{3}}}{\mathrm{4}}−\frac{\mathrm{3}\:\mathrm{ln}\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)}{\mathrm{8}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{60}}\left[\mathrm{256}\pi−\mathrm{66}\sqrt{\mathrm{3}}−\mathrm{9}\:\mathrm{ln}\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\right] \\ $$$$\approx\mathrm{11}.\mathrm{301} \\ $$$$ \\ $$$$\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\theta}{\mathrm{sin}^{\mathrm{5}} \:\theta} \\ $$$$=\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}\:\theta{d}\theta}{\mathrm{sin}^{\mathrm{6}} \:\theta} \\ $$$$=−\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\left(\mathrm{cos}\:\theta\right)}{\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\theta\right)^{\mathrm{3}} } \\ $$$$=\int_{\mathrm{0}} ^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \frac{{du}}{\left(\mathrm{1}−{u}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$$...... \\ $$$$=\frac{\mathrm{3}}{\mathrm{8}}\mathrm{ln}\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)+\frac{\mathrm{11}\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$
Commented by mr W last updated on 12/Apr/20
$$\begin{cases}{\theta=\frac{\pi}{\mathrm{6}}}\\{\theta=\frac{\mathrm{5}\pi}{\mathrm{6}}}\end{cases} \\ $$$${equal}\:{to}\:\mathrm{2}\:{times}\:\theta\:{from}\:\frac{\pi}{\mathrm{6}}\:{to}\:\frac{\pi}{\mathrm{2}}\:{due}\:{to} \\ $$$${symmetry} \\ $$
Commented by john santu last updated on 12/Apr/20
$${sir},\:{if}\:{r}\:=\:\mathrm{2}\:,\:{y}\:=\:\mathrm{1} \\ $$$${then}\:\mathrm{2}\:\mathrm{sin}\:\theta\:=\:\mathrm{1}\:\begin{cases}{\theta=\frac{\pi}{\mathrm{6}}}\\{\theta=\frac{\mathrm{5}\pi}{\mathrm{6}}}\end{cases} \\ $$
Commented by mr W last updated on 12/Apr/20
$${our}\:{results}\:{are}\:{the}\:{same}. \\ $$
Commented by john santu last updated on 12/Apr/20
$${oo}\:{yes}\:{sir}.\:{thank}\:{you} \\ $$
Commented by jagoll last updated on 12/Apr/20
$${thank}\:{you}\:{mr}\:{W} \\ $$