Evaluate: ∫_R ∫ ((xy)/(√(1−y^2 ))) dx dy where the region of integration is the positive quadrant of the circle x^2 +y^2 =1.


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Question Number 92805 by niroj last updated on 09/May/20

$Evaluate :$ $\int_{R} \int \frac{xy}{\sqrt{1 - y^{2}}} dx dy where the region of integration is the$ $positive quadrant of the circle x^{2} + y^{2} = 1 .$
	Answered by mr W last updated on 09/May/20
	$=∫_0 ^(π/2) ∫_0 ^1 ((r^2 cos θ sin θ)/(√(1−r^2 sin^2 θ)))rdrdθ =(1/2)∫_0 ^(π/2) cos θ sin θ(∫_0 ^1 (r^2 /(√(1−r^2 sin^2 θ)))dr^2 )dθ =(1/2)∫_0 ^(π/2) cos θ sin θ(∫_0 ^1 (u/(√(1−u sin^2 θ)))du)dθ =(1/2)∫_0 ^(π/2) cos θ sin θ(1/(sin^2 θ)){∫_0 ^1 [−(√(1−u sin^2 θ))+(1/(√(1−u sin^2 θ)))]du}dθ =(1/2)∫_0 ^(π/2) ((cos θ)/(sin θ)){[(2/(3 sin^2 θ))(1−u sin^2 θ)^(3/2) −(2/(sin^2 θ))(√(1−u sin^2 θ))]_0 ^1 }dθ =(1/2)∫_0 ^(π/2) ((cos θ)/(sin θ)){[(2/(3 sin^2 θ))(1−sin^2 θ)^(3/2) −(2/(3 sin^2 θ))−(2/(sin^2 θ))(√(1−sin^2 θ))+(2/(sin^2 θ))]}dθ =(1/3)∫_0 ^(π/2) ((cos θ)/(sin θ))[((cos^3 θ−3cos θ+2)/(sin^2 θ))]dθ =(1/3)∫_(π/2) ^0 (((cos^3 θ−3cos θ+2)cos θ)/((1−cos^2 θ)^2 ))d(cos θ) =(1/3)∫_0 ^1 (((t^3 −3t+2)t)/((1−t^2 )^2 ))dt =(1/3)∫_0 ^1 (((t+2)t)/((t+1)^2 ))dt =(1/3)∫_0 ^1 [1−(1/((t+1)^2 ))]dt =(1/3)[t+(1/(t+1))]_0 ^1 =(1/3)(1+(1/2)−1) =(1/6)$
	$= \int_{0}^{\frac{π}{2}} \int_{0}^{1} \frac{r^{2} \cos θ \sin θ}{\sqrt{1 - r^{2} \sin^{2} θ}} r d r d θ$ $= \frac{1}{2} \int_{0}^{\frac{π}{2}} \cos θ \sin θ (\int_{0}^{1} \frac{r^{2}}{\sqrt{1 - r^{2} \sin^{2} θ}} {d r}^{2}) d θ$ $= \frac{1}{2} \int_{0}^{\frac{π}{2}} \cos θ \sin θ (\int_{0}^{1} \frac{u}{\sqrt{1 - u \sin^{2} θ}} d u) d θ$ $= \frac{1}{2} \int_{0}^{\frac{π}{2}} \cos θ \sin θ \frac{1}{\sin^{2} θ} {\int_{0}^{1} [- \sqrt{1 - u \sin^{2} θ} + \frac{1}{\sqrt{1 - u \sin^{2} θ}}] d u} d θ$ $= \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{\cos θ}{\sin θ} {{[\frac{2}{3 \sin^{2} θ} {(1 - u \sin^{2} θ)}^{\frac{3}{2}} - \frac{2}{\sin^{2} θ} \sqrt{1 - u \sin^{2} θ}]}_{0}^{1}} d θ$ $= \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{\cos θ}{\sin θ} {[\frac{2}{3 \sin^{2} θ} {(1 - \sin^{2} θ)}^{\frac{3}{2}} - \frac{2}{3 \sin^{2} θ} - \frac{2}{\sin^{2} θ} \sqrt{1 - \sin^{2} θ} + \frac{2}{\sin^{2} θ}]} d θ$ $= \frac{1}{3} \int_{0}^{\frac{π}{2}} \frac{\cos θ}{\sin θ} [\frac{\cos^{3} θ - 3 \cos θ + 2}{\sin^{2} θ}] d θ$ $= \frac{1}{3} \int_{\frac{π}{2}}^{0} \frac{(\cos^{3} θ - 3 \cos θ + 2) \cos θ}{{(1 - \cos^{2} θ)}^{2}} d (\cos θ)$ $= \frac{1}{3} \int_{0}^{1} \frac{(t^{3} - 3 t + 2) t}{{(1 - t^{2})}^{2}} d t$ $= \frac{1}{3} \int_{0}^{1} \frac{(t + 2) t}{{(t + 1)}^{2}} d t$ $= \frac{1}{3} \int_{0}^{1} [1 - \frac{1}{{(t + 1)}^{2}}] d t$ $= \frac{1}{3} {[t + \frac{1}{t + 1}]}_{0}^{1}$ $= \frac{1}{3} (1 + \frac{1}{2} - 1)$ $= \frac{1}{6}$
	Commented by niroj last updated on 09/May/20
	Thank you Mr. W It is aspected result ��
	Commented by Ar Brandon last updated on 09/May/20
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