Question Number 72466 by Learner-123 last updated on 29/Oct/19
$${Prove}\:{that}\:\int_{\mathrm{0}} ^{\:\mathrm{2}} \int_{−\sqrt{\mathrm{1}−\left({y}−\mathrm{1}\right)^{\mathrm{2}} }} ^{\:\:\mathrm{0}\:} \:{xy}^{\mathrm{2}} {dxdy}\:=\:−\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\boldsymbol{{after}}\:\mathrm{changing}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{to}\:\boldsymbol{\mathrm{polar}}\:\boldsymbol{\mathrm{form}}. \\ $$
Commented by Abdo msup. last updated on 30/Oct/19
![I=∫_0 ^2 ( ∫_(−(√(1−(y−1)^2 ))) ^0 xdx)y^2 dy but ∫_(−(√(1−(y−1)^2 ))) ^0 xdx =[(x^2 /2)]_(−(√(1−(y−1)^2 ))) ^0 =−((1−(y−1)^2 )/2) =−((1−y^2 +2y−1)/2) =((y^2 −2y)/2) ⇒ I =(1/2)∫_0 ^2 (y^2 −2y)y^2 dy =(1/2)∫_0 ^2 (y^4 −2y^3 )dy =(1/2)[(y^5 /5)−(1/2)y^4 ]_0 ^2 =(1/2){((32)/5)−8}=(1/2)×((−8)/5) =−(4/5) without use of polar coordinates.](Q72556.png)
$${I}=\int_{\mathrm{0}} ^{\mathrm{2}} \left(\:\int_{−\sqrt{\mathrm{1}−\left({y}−\mathrm{1}\right)^{\mathrm{2}} }} ^{\mathrm{0}} \:\:{xdx}\right){y}^{\mathrm{2}} {dy}\:{but} \\ $$$$\int_{−\sqrt{\mathrm{1}−\left({y}−\mathrm{1}\right)^{\mathrm{2}} }} ^{\mathrm{0}} {xdx}\:=\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{−\sqrt{\mathrm{1}−\left({y}−\mathrm{1}\right)^{\mathrm{2}} }} ^{\mathrm{0}} =−\frac{\mathrm{1}−\left({y}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$$=−\frac{\mathrm{1}−{y}^{\mathrm{2}} +\mathrm{2}{y}−\mathrm{1}}{\mathrm{2}}\:=\frac{{y}^{\mathrm{2}} −\mathrm{2}{y}}{\mathrm{2}}\:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}} \left({y}^{\mathrm{2}} −\mathrm{2}{y}\right){y}^{\mathrm{2}} {dy}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}} \left({y}^{\mathrm{4}} −\mathrm{2}{y}^{\mathrm{3}} \right){dy} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{y}^{\mathrm{5}} }{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{2}}{y}^{\mathrm{4}} \right]_{\mathrm{0}} ^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{32}}{\mathrm{5}}−\mathrm{8}\right\}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{−\mathrm{8}}{\mathrm{5}}\:=−\frac{\mathrm{4}}{\mathrm{5}} \\ $$$${without}\:{use}\:{of}\:{polar}\:{coordinates}. \\ $$$$ \\ $$
Commented by Learner-123 last updated on 30/Oct/19
$${thanks}\:{sir}. \\ $$
Answered by mind is power last updated on 29/Oct/19
![x=rcos(θ) y=rsin(θ) dxdy=rdrdθ 0≥x≥−(√(1−(y−1)))^2 0≤x^2 ≤1−(y−1)^2 =2y−y^2 ⇒0≤x^2 +y^2 ≤2y 0≤r^2 ≤2rsin(θ)⇒0≤r≤2sin(θ) 0≤y≤2 ⇒0≤rsin(θ)≤2⇒sin(θ)≥0⇒θ∈[0,π] but x≤0⇒cos(θ)≤0⇒θ∈[(π/2),2] our integral ⇔∫_((π/2) ) ^π ∫_0 ^(2sin(θ)) .rcos(θ).r^2 sin^2 (θ).rdrdθ =∫_(π/2) ^π ∫_0 ^(2sin(θ)) .r^4 cos(θ)sin^2 (θ)drdθ](Q72469.png)
$$\mathrm{x}=\mathrm{rcos}\left(\theta\right) \\ $$$$\mathrm{y}=\mathrm{rsin}\left(\theta\right) \\ $$$$\mathrm{dxdy}=\mathrm{rdrd}\theta \\ $$$$\left.\:\:\:\:\:\:\:\mathrm{0}\geqslant\mathrm{x}\geqslant−\sqrt{\mathrm{1}−\left(\mathrm{y}−\mathrm{1}\right.}\right)^{\mathrm{2}} \\ $$$$\mathrm{0}\leqslant\mathrm{x}^{\mathrm{2}} \leqslant\mathrm{1}−\left(\mathrm{y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{2y}−\mathrm{y}^{\mathrm{2}} \Rightarrow\mathrm{0}\leqslant\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \leqslant\mathrm{2y} \\ $$$$\mathrm{0}\leqslant\mathrm{r}^{\mathrm{2}} \leqslant\mathrm{2rsin}\left(\theta\right)\Rightarrow\mathrm{0}\leqslant\mathrm{r}\leqslant\mathrm{2sin}\left(\theta\right) \\ $$$$\mathrm{0}\leqslant\mathrm{y}\leqslant\mathrm{2} \\ $$$$\Rightarrow\mathrm{0}\leqslant\mathrm{rsin}\left(\theta\right)\leqslant\mathrm{2}\Rightarrow\mathrm{sin}\left(\theta\right)\geqslant\mathrm{0}\Rightarrow\theta\in\left[\mathrm{0},\pi\right] \\ $$$$\mathrm{but}\:\mathrm{x}\leqslant\mathrm{0}\Rightarrow\mathrm{cos}\left(\theta\right)\leqslant\mathrm{0}\Rightarrow\theta\in\left[\frac{\pi}{\mathrm{2}},\mathrm{2}\right] \\ $$$$\mathrm{our}\:\mathrm{integral}\:\Leftrightarrow\int_{\frac{\pi}{\mathrm{2}}\:} ^{\pi} \int_{\mathrm{0}} ^{\mathrm{2sin}\left(\theta\right)} .\mathrm{rcos}\left(\theta\right).\mathrm{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\theta\right).\mathrm{rdrd}\theta \\ $$$$=\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \int_{\mathrm{0}} ^{\mathrm{2sin}\left(\theta\right)} .\mathrm{r}^{\mathrm{4}} \mathrm{cos}\left(\theta\right)\mathrm{sin}^{\mathrm{2}} \left(\theta\right)\mathrm{drd}\theta \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by Learner-123 last updated on 29/Oct/19
$${One}\:{more}\:{Ques}.\::: \\ $$$$\:\int_{−{a}} ^{\:{a}} \int_{−\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }} ^{\:\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }} \:{dydx}. \\ $$
Commented by Learner-123 last updated on 29/Oct/19
$${thank}\:{you}\:{sir}! \\ $$
Commented by mind is power last updated on 29/Oct/19
$$\mathrm{y}'\mathrm{re}\:\mathrm{Welcom} \\ $$
Commented by mind is power last updated on 29/Oct/19
![x=rcos(b) y=rsin(b) we have −a<x<a −(√(a^2 −x^2 ))<y<(√(a^2 −x^2 ))⇒0<x^2 +y^2 ≤a^2 ⇒0<R<a b∈[0,2π] ∫_0 ^(2π) ∫_0 ^a rdrdb =2π.((a^2 /2))=πa^2](Q72485.png)
$$\mathrm{x}=\mathrm{rcos}\left(\mathrm{b}\right) \\ $$$$\mathrm{y}=\mathrm{rsin}\left(\mathrm{b}\right) \\ $$$$\mathrm{we}\:\mathrm{have}\:−\mathrm{a}<\mathrm{x}<\mathrm{a} \\ $$$$−\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} }<\mathrm{y}<\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} }\Rightarrow\mathrm{0}<\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \leqslant\mathrm{a}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{0}<\mathrm{R}<\mathrm{a} \\ $$$$\mathrm{b}\in\left[\mathrm{0},\mathrm{2}\pi\right] \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \int_{\mathrm{0}} ^{\mathrm{a}} \mathrm{rdrdb} \\ $$$$=\mathrm{2}\pi.\left(\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}\right)=\pi\mathrm{a}^{\mathrm{2}} \\ $$
Commented by Learner-123 last updated on 29/Oct/19
$${thanks}\:{again}\:,{sir}! \\ $$
Commented by mind is power last updated on 29/Oct/19
$$\mathrm{most}\:\mathrm{Welcom} \\ $$