Evaluate : the Integral ∫_(-(π/2)) ^(π/2) ∫_0 ^(3 cos θ) r^2 sin^2 θ. dr dθ


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Question Number 37912 by gunawan last updated on 19/Jun/18

$Evaluate : the Integral$ $\int_{- \frac{π}{2}}^{\frac{π}{2}} \int_{0}^{3 \cos θ} r^{2} \sin^{2} θ . d r d θ$
Commented by math khazana by abdo last updated on 19/Jun/18
$I = ∫_(−(π/2)) ^(π/2) A(θ)sin^2 θdθ with A(θ) = ∫_0 ^(3cosθ) r^2 dr =(1/3)[r^3 ]_0 ^(3cosθ) =(1/3) 27 cos^3 θ =9 cos^3 θ ⇒I =9∫_(−(π/2)) ^(π/2) sin^2 θ cos^3 θ dθ =_(sinθ =t) ∫_(−1) ^1 t^2 (1−t^2 )(√(1−t^2 )) (dt/(√(1−t^2 ))) =∫_(−1) ^1 (t^2 −t^4 )dt=2[ (t^3 /3) −(t^5 /5)]_0 ^1 =2 { (1/3) −(1/5)}=2(2/(15)) =(4/(15)) ⇒ I =9 .(4/(15)) = ((12)/5) I =((12)/5) .$
$I = \int_{- \frac{π}{2}}^{\frac{π}{2}} A (θ) {s i n}^{2} θ d θ w i t h$ $A (θ) = \int_{0}^{3 c o s θ} r^{2} d r = \frac{1}{3} {[r^{3}]}_{0}^{3 c o s θ} = \frac{1}{3} 27 {c o s}^{3} θ$ $= 9 {c o s}^{3} θ \Rightarrow I = 9 \int_{- \frac{π}{2}}^{\frac{π}{2}} {s i n}^{2} θ {c o s}^{3} θ d θ$ $=_{s i n θ = t} \int_{- 1}^{1} t^{2} (1 - t^{2}) \sqrt{1 - t^{2}} \frac{d t}{\sqrt{1 - t^{2}}}$ $= \int_{- 1}^{1} (t^{2} - t^{4}) d t = 2 {[\frac{t^{3}}{3} - \frac{t^{5}}{5}]}_{0}^{1}$ $= 2 {\frac{1}{3} - \frac{1}{5}} = 2 \frac{2}{15} = \frac{4}{15} \Rightarrow I = 9 . \frac{4}{15} = \frac{12}{5}$ $I = \frac{12}{5} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jun/18
	$∫_((−Π)/2) ^(Π/2) sin^2 θ×∣(r^3 /3)∣_0 ^(3cosθ) dθ (1/3)∫_((−Π)/2) ^(Π/2) sin^2 θ×27cos^3 θdθ 9∫_(−(Π/2)) ^(Π/2) sin^2 θ.(1−sin^2 θ)cosθdθ t=sinθ dt=cosθdθ 9∫_(−1) ^1 (t^2 −t^4 )dt 9∣(t^3 /3)−(t^5 /5)∣_(−1) ^1 9{((1/3)−(1/5))−(((−1)/3)−((−1)/5))} 9{(1/3)−(1/5)+(1/3)−(1/5)} 18((1/3)−(1/5))=18×(2/(15))=((12)/5)$
	$\int_{\frac{- Π}{2}}^{\frac{Π}{2}} {s i n}^{2} θ \times ∣ \frac{r^{3}}{3} ∣_{0}^{3 c o s θ} d θ$ $\frac{1}{3} \int_{\frac{- Π}{2}}^{\frac{Π}{2}} {s i n}^{2} θ \times 27 {c o s}^{3} θ d θ$ $9 \int_{- \frac{Π}{2}}^{\frac{Π}{2}} {s i n}^{2} θ . (1 - {s i n}^{2} θ) c o s θ d θ$ $t = s i n θ d t = c o s θ d θ$ $9 \int_{- 1}^{1} (t^{2} - t^{4}) d t$ $9 ∣ \frac{t^{3}}{3} - \frac{t^{5}}{5} ∣_{- 1}^{1}$ $9 {(\frac{1}{3} - \frac{1}{5}) - (\frac{- 1}{3} - \frac{- 1}{5})}$ $9 {\frac{1}{3} - \frac{1}{5} + \frac{1}{3} - \frac{1}{5}}$ $18 (\frac{1}{3} - \frac{1}{5}) = 18 \times \frac{2}{15} = \frac{12}{5}$
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