Question Number 37912 by gunawan last updated on 19/Jun/18
$$\mathrm{Evaluate}\::\:\mathrm{the}\:\mathrm{Integral} \\ $$$$\int_{-\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\mathrm{3}\:\mathrm{cos}\:\theta} {r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \theta.\:{dr}\:{d}\theta\: \\ $$
Commented by math khazana by abdo last updated on 19/Jun/18
![I = ∫_(−(π/2)) ^(π/2) A(θ)sin^2 θdθ with A(θ) = ∫_0 ^(3cosθ) r^2 dr =(1/3)[r^3 ]_0 ^(3cosθ) =(1/3) 27 cos^3 θ =9 cos^3 θ ⇒I =9∫_(−(π/2)) ^(π/2) sin^2 θ cos^3 θ dθ =_(sinθ =t) ∫_(−1) ^1 t^2 (1−t^2 )(√(1−t^2 )) (dt/( (√(1−t^2 )))) =∫_(−1) ^1 (t^2 −t^4 )dt=2[ (t^3 /3) −(t^5 /5)]_0 ^1 =2 { (1/3) −(1/5)}=2(2/(15)) =(4/(15)) ⇒ I =9 .(4/(15)) = ((12)/5) I =((12)/5) .](https://www.tinkutara.com/question/Q37927.png)
$${I}\:=\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} {A}\left(\theta\right){sin}^{\mathrm{2}} \theta{d}\theta\:{with} \\ $$$${A}\left(\theta\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{3}{cos}\theta} \:{r}^{\mathrm{2}} {dr}\:=\frac{\mathrm{1}}{\mathrm{3}}\left[{r}^{\mathrm{3}} \right]_{\mathrm{0}} ^{\mathrm{3}{cos}\theta} =\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{27}\:{cos}^{\mathrm{3}} \theta \\ $$$$=\mathrm{9}\:{cos}^{\mathrm{3}} \theta\:\Rightarrow{I}\:=\mathrm{9}\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}^{\mathrm{2}} \theta\:{cos}^{\mathrm{3}} \theta\:{d}\theta \\ $$$$=_{{sin}\theta\:={t}} \:\:\:\:\int_{−\mathrm{1}} ^{\mathrm{1}} \:{t}^{\mathrm{2}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:\frac{{dt}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }} \\ $$$$=\int_{−\mathrm{1}} ^{\mathrm{1}} \:\left({t}^{\mathrm{2}} \:−{t}^{\mathrm{4}} \right){dt}=\mathrm{2}\left[\:\frac{{t}^{\mathrm{3}} }{\mathrm{3}}\:−\frac{{t}^{\mathrm{5}} }{\mathrm{5}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\mathrm{2}\:\left\{\:\frac{\mathrm{1}}{\mathrm{3}}\:−\frac{\mathrm{1}}{\mathrm{5}}\right\}=\mathrm{2}\frac{\mathrm{2}}{\mathrm{15}}\:=\frac{\mathrm{4}}{\mathrm{15}}\:\Rightarrow\:{I}\:=\mathrm{9}\:.\frac{\mathrm{4}}{\mathrm{15}}\:=\:\frac{\mathrm{12}}{\mathrm{5}} \\ $$$${I}\:=\frac{\mathrm{12}}{\mathrm{5}}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jun/18
$$\int_{\frac{−\Pi}{\mathrm{2}}} ^{\frac{\Pi}{\mathrm{2}}} \:{sin}^{\mathrm{2}} \theta×\mid\frac{{r}^{\mathrm{3}} }{\mathrm{3}}\mid_{\mathrm{0}} ^{\mathrm{3}{cos}\theta} {d}\theta \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\int_{\frac{−\Pi}{\mathrm{2}}} ^{\frac{\Pi}{\mathrm{2}}} \:{sin}^{\mathrm{2}} \theta×\mathrm{27}{cos}^{\mathrm{3}} \theta{d}\theta \\ $$$$\mathrm{9}\int_{−\frac{\Pi}{\mathrm{2}}} ^{\frac{\Pi}{\mathrm{2}}} \:{sin}^{\mathrm{2}} \theta.\left(\mathrm{1}−{sin}^{\mathrm{2}} \theta\right){cos}\theta{d}\theta \\ $$$${t}={sin}\theta\:\:{dt}={cos}\theta{d}\theta \\ $$$$\mathrm{9}\int_{−\mathrm{1}} ^{\mathrm{1}} \left({t}^{\mathrm{2}} −{t}^{\mathrm{4}} \right){dt} \\ $$$$\mathrm{9}\mid\frac{{t}^{\mathrm{3}} }{\mathrm{3}}−\frac{{t}^{\mathrm{5}} }{\mathrm{5}}\mid_{−\mathrm{1}} ^{\mathrm{1}} \\ $$$$\mathrm{9}\left\{\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{5}}\right)−\left(\frac{−\mathrm{1}}{\mathrm{3}}−\frac{−\mathrm{1}}{\mathrm{5}}\right)\right\} \\ $$$$\mathrm{9}\left\{\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{5}}\right\} \\ $$$$\mathrm{18}\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{5}}\right)=\mathrm{18}×\frac{\mathrm{2}}{\mathrm{15}}=\frac{\mathrm{12}}{\mathrm{5}} \\ $$$$ \\ $$