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Question Number 36191 by prof Abdo imad last updated on 30/May/18
let D = {(x,y)∈R^2  /x>0 ,y>0,x+y<1}  1) calculate ∫∫_D   ((xy)/(x^2  +y^2 ))dxdy  2) let a>0 ,b>0 calculate ∫∫_D  a^x b^y dxdy
$${let}\:{D}\:=\:\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} \:/{x}>\mathrm{0}\:,{y}>\mathrm{0},{x}+{y}<\mathrm{1}\right\} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\int\int_{{D}} \:\:\frac{{xy}}{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }{dxdy} \\ $$$$\left.\mathrm{2}\right)\:{let}\:{a}>\mathrm{0}\:,{b}>\mathrm{0}\:{calculate}\:\int\int_{{D}} \:{a}^{{x}} {b}^{{y}} {dxdy} \\ $$
Commented by maxmathsup by imad last updated on 26/Aug/18
polar coordinates changement x =rcosθ  and y =rsinθ    x>0 and y>0 ⇒ 0<θ<(π/2)  and x+y <1 ⇒ r(cosθ +sinθ)<1 ⇒  0<r<(1/(cosθ +sinθ))  ∫∫_D ((xy)/(x^2  +y^2 ))dxdy  = ∫_0 ^(π/2)   (∫_0 ^(1/(cosθ+sinθ)) ((r^2 cosθsinθ)/r^2 ) rdr)dθ  =∫_0 ^(π/2)   ( ∫_0 ^(1/(cosθ +sinθ)) rdr)cosθ sinθ dθ   but  ∫_0 ^(1/(cosθ +sinθ))  rdr =[(r^2 /2)]_0 ^(1/(cosθ +sinθ))  =(1/(2(cosθ +sinθ)^2 )) =(1/(2(1+2sinθ cosθ))) ⇒  I  = ∫_0 ^(π/2)     ((cosθ sinθ)/(2(1+2sinθ cosθ)))dθ =(1/4) ∫_0 ^(π/2)  ((sin(2θ))/(1+sin(2θ)))dθ  =_(2θ =t)     (1/4) ∫_0 ^π     ((sint)/(1+sint)) (dt/2) =(1/8) ∫_0 ^π   ((1+sint −1)/(1+sint))dt  =(π/8)  −(1/8) ∫_0 ^π     (dt/(1+sint))  chang. tan((t/2)) =u give  ∫_0 ^π     (dt/(1+sint))  = ∫_0 ^∞      (1/(1+((2u)/(1+u^2 )))) ((2du)/(1+u^2 )) =2 ∫_0 ^∞    (du/(1+u^2  +2u))  =2 ∫_0 ^∞      (du/((u+1)^2 )) =2[−(1/(u+1))]_0 ^(+∞)  = 2  ⇒ I =(π/8) −(1/4) .
$${polar}\:{coordinates}\:{changement}\:{x}\:={rcos}\theta\:\:{and}\:{y}\:={rsin}\theta\:\: \\ $$$${x}>\mathrm{0}\:{and}\:{y}>\mathrm{0}\:\Rightarrow\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}}\:\:{and}\:{x}+{y}\:<\mathrm{1}\:\Rightarrow\:{r}\left({cos}\theta\:+{sin}\theta\right)<\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{0}<{r}<\frac{\mathrm{1}}{{cos}\theta\:+{sin}\theta} \\ $$$$\int\int_{{D}} \frac{{xy}}{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }{dxdy}\:\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\left(\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{{cos}\theta+{sin}\theta}} \frac{{r}^{\mathrm{2}} {cos}\theta{sin}\theta}{{r}^{\mathrm{2}} }\:{rdr}\right){d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\left(\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{{cos}\theta\:+{sin}\theta}} {rdr}\right){cos}\theta\:{sin}\theta\:{d}\theta\:\:\:{but} \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{{cos}\theta\:+{sin}\theta}} \:{rdr}\:=\left[\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{{cos}\theta\:+{sin}\theta}} \:=\frac{\mathrm{1}}{\mathrm{2}\left({cos}\theta\:+{sin}\theta\right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+\mathrm{2}{sin}\theta\:{cos}\theta\right)}\:\Rightarrow \\ $$$${I}\:\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{cos}\theta\:{sin}\theta}{\mathrm{2}\left(\mathrm{1}+\mathrm{2}{sin}\theta\:{cos}\theta\right)}{d}\theta\:=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{sin}\left(\mathrm{2}\theta\right)}{\mathrm{1}+{sin}\left(\mathrm{2}\theta\right)}{d}\theta \\ $$$$=_{\mathrm{2}\theta\:={t}} \:\:\:\:\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{sint}}{\mathrm{1}+{sint}}\:\frac{{dt}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{8}}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{\mathrm{1}+{sint}\:−\mathrm{1}}{\mathrm{1}+{sint}}{dt} \\ $$$$=\frac{\pi}{\mathrm{8}}\:\:−\frac{\mathrm{1}}{\mathrm{8}}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{dt}}{\mathrm{1}+{sint}}\:\:{chang}.\:{tan}\left(\frac{{t}}{\mathrm{2}}\right)\:={u}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{dt}}{\mathrm{1}+{sint}}\:\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} \:+\mathrm{2}{u}} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{du}}{\left({u}+\mathrm{1}\right)^{\mathrm{2}} }\:=\mathrm{2}\left[−\frac{\mathrm{1}}{{u}+\mathrm{1}}\right]_{\mathrm{0}} ^{+\infty} \:=\:\mathrm{2}\:\:\Rightarrow\:{I}\:=\frac{\pi}{\mathrm{8}}\:−\frac{\mathrm{1}}{\mathrm{4}}\:. \\ $$$$ \\ $$

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