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Question Number 56011 by mr W last updated on 08/Mar/19
Commented by mr W last updated on 08/Mar/19
$${Point}\:{A}\:{is}\:{choosen}\:{at}\:{random}\:{in}\:{a} \\ $$$${square}\:{with}\:{side}\:{length}\:{a},\:{see}\:{figure}. \\ $$$${What}\:{is}\:{its}\:{average}\:{distance}\:{to}\:{the} \\ $$$${origin}? \\ $$$${Take}\:{P}\left({h},{k}\right)=\left(\mathrm{0},\mathrm{0}\right)\:{at}\:{first}. \\ $$
Answered by ajfour last updated on 08/Mar/19
![For h>k (If h<k we will swap h with k) a^2 r_(avg) ^ =∫_(√(h^2 +k^2 )) ^( (√(h^2 +(k+a)^2 ))) [cos^(−1) ((h/r))−sin^(−1) ((k/r))]r^2 dr +∫_(√(h^2 +(k+a)^2 )) ^( (√((h+a)^2 +k^2 ))) [sin^(−1) (((k+a)/r))−sin^(−1) ((k/r))]r^2 dr +∫_(√((h+a)^2 +k^2 )) ^( (√((h+a)^2 +(k+a)^2 ))) [sin^(−1) (((k+a)/r))−cos^(−1) (((h+a)/r))]r^2 dr . If (h,k)≡(0,0)_(−) a^2 r_(avg) =(π/2)∫_0 ^( a) r^2 dr+∫_a ^( a(√2)) r^2 [sin^(−1) ((a/r))−cos^(−1) ((a/r))]dr =((πa^3 )/6)+∫_a ^( a(√2)) r^2 [(π/2)−2cos^(−1) ((a/r))]dr =((πa^3 )/6)+((πa^3 )/6)(2(√2)−1)−2I =((2(√2)πa^3 )/6)−2I I=∫_a ^( a(√2)) r^2 cos^(−1) ((a/r))dr let cos θ=(a/r) ⇒ r=asec θ and dr=asec θtan θdθ ⇒ I=∫_0 ^( π/4) a^3 θsec^3 θtan θdθ = a^3 [θ∫_0 ^( π/4) sec^2 θ(sec θtan θdθ) −(1/3)∫_0 ^( π/4) sec^3 θdθ ] = a^3 [(π/4)×((2(√2))/3)−(1/3)∫_0 ^( π/4) sec^3 θdθ] ∫sec^3 θdθ = sec θtan θ−∫sec θtan^2 θdθ 2∫sec^3 θdθ = sec θtan θ+∫sec θdθ 2∫sec^3 θdθ=sec θtan θ+ln ∣sec θ+tan θ∣+c ⇒ 2∫_0 ^( π/4) sec^3 θ=(√2)+ln ((√2)+1) a^2 r_(avg) =((2(√2)πa^3 )/6)−2a^3 [(π/4)×((2(√2))/3)−(1/3)∫_0 ^( π/4) sec^3 θdθ] a^2 r_(avg) =((2(√2)πa^3 )/6)−2a^3 [(π/4)×((2(√2))/3)−(1/6)((√2)+ln ((√2)+1))] ⇒ r_(avg) =(a/3)[(√2)+ln ((√2)+1)] ⇒ r_(avg) ≈ 0.765a ■](Q56020.png)
$$\:\:\:{For}\:{h}>{k}\:\:\left({If}\:\:{h}<{k}\:\:{we}\:{will}\:{swap}\:{h}\:{with}\:{k}\right) \\ $$$$\:\:\:{a}^{\mathrm{2}} \bar {{r}}_{{avg}} =\int_{\sqrt{{h}^{\mathrm{2}} +{k}^{\mathrm{2}} }} ^{\:\sqrt{{h}^{\mathrm{2}} +\left({k}+{a}\right)^{\mathrm{2}} }} \left[\mathrm{cos}^{−\mathrm{1}} \left(\frac{{h}}{{r}}\right)−\mathrm{sin}^{−\mathrm{1}} \left(\frac{{k}}{{r}}\right)\right]{r}^{\mathrm{2}} {dr} \\ $$$$\:\:+\int_{\sqrt{{h}^{\mathrm{2}} +\left({k}+{a}\right)^{\mathrm{2}} }} ^{\:\sqrt{\left({h}+{a}\right)^{\mathrm{2}} +{k}^{\mathrm{2}} }} \left[\mathrm{sin}^{−\mathrm{1}} \left(\frac{{k}+{a}}{{r}}\right)−\mathrm{sin}^{−\mathrm{1}} \left(\frac{{k}}{{r}}\right)\right]{r}^{\mathrm{2}} {dr} \\ $$$$\:+\int_{\sqrt{\left({h}+{a}\right)^{\mathrm{2}} +{k}^{\mathrm{2}} }} ^{\:\sqrt{\left({h}+{a}\right)^{\mathrm{2}} +\left({k}+{a}\right)^{\mathrm{2}} }} \left[\mathrm{sin}^{−\mathrm{1}} \left(\frac{{k}+{a}}{{r}}\right)−\mathrm{cos}^{−\mathrm{1}} \left(\frac{{h}+{a}}{{r}}\right)\right]{r}^{\mathrm{2}} {dr}\:. \\ $$$$ \\ $$$$\underset{−} {{If}\:\left({h},{k}\right)\equiv\left(\mathrm{0},\mathrm{0}\right)} \\ $$$${a}^{\mathrm{2}} {r}_{{avg}} =\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\:{a}} {r}^{\mathrm{2}} {dr}+\int_{{a}} ^{\:{a}\sqrt{\mathrm{2}}} {r}^{\mathrm{2}} \left[\mathrm{sin}^{−\mathrm{1}} \left(\frac{{a}}{{r}}\right)−\mathrm{cos}^{−\mathrm{1}} \left(\frac{{a}}{{r}}\right)\right]{dr} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\pi{a}^{\mathrm{3}} }{\mathrm{6}}+\int_{{a}} ^{\:{a}\sqrt{\mathrm{2}}} {r}^{\mathrm{2}} \left[\frac{\pi}{\mathrm{2}}−\mathrm{2cos}^{−\mathrm{1}} \left(\frac{{a}}{{r}}\right)\right]{dr} \\ $$$$\:\:\:=\frac{\pi{a}^{\mathrm{3}} }{\mathrm{6}}+\frac{\pi{a}^{\mathrm{3}} }{\mathrm{6}}\left(\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}\right)−\mathrm{2}{I} \\ $$$$\:\:\:=\frac{\mathrm{2}\sqrt{\mathrm{2}}\pi{a}^{\mathrm{3}} }{\mathrm{6}}−\mathrm{2}{I} \\ $$$${I}=\int_{{a}} ^{\:{a}\sqrt{\mathrm{2}}} {r}^{\mathrm{2}} \mathrm{cos}^{−\mathrm{1}} \left(\frac{{a}}{{r}}\right){dr} \\ $$$${let}\:\:\:\:\mathrm{cos}\:\theta=\frac{{a}}{{r}}\:\Rightarrow\:{r}={a}\mathrm{sec}\:\theta \\ $$$${and}\:\:\:{dr}={a}\mathrm{sec}\:\theta\mathrm{tan}\:\theta{d}\theta \\ $$$$\Rightarrow\:{I}=\int_{\mathrm{0}} ^{\:\pi/\mathrm{4}} {a}^{\mathrm{3}} \theta\mathrm{sec}\:^{\mathrm{3}} \theta\mathrm{tan}\:\theta{d}\theta \\ $$$$\:=\:{a}^{\mathrm{3}} \left[\theta\int_{\mathrm{0}} ^{\:\pi/\mathrm{4}} \mathrm{sec}\:^{\mathrm{2}} \theta\left(\mathrm{sec}\:\theta\mathrm{tan}\:\theta{d}\theta\right)\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\:\pi/\mathrm{4}} \mathrm{sec}\:^{\mathrm{3}} \theta{d}\theta\:\right] \\ $$$$\:\:\:\:\:=\:{a}^{\mathrm{3}} \left[\frac{\pi}{\mathrm{4}}×\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\:\pi/\mathrm{4}} \mathrm{sec}\:^{\mathrm{3}} \theta{d}\theta\right] \\ $$$$\int\mathrm{sec}\:^{\mathrm{3}} \theta{d}\theta\:=\:\mathrm{sec}\:\theta\mathrm{tan}\:\theta−\int\mathrm{sec}\:\theta\mathrm{tan}\:^{\mathrm{2}} \theta{d}\theta \\ $$$$\mathrm{2}\int\mathrm{sec}\:^{\mathrm{3}} \theta{d}\theta\:=\:\mathrm{sec}\:\theta\mathrm{tan}\:\theta+\int\mathrm{sec}\:\theta{d}\theta \\ $$$$\mathrm{2}\int\mathrm{sec}\:^{\mathrm{3}} \theta{d}\theta=\mathrm{sec}\:\theta\mathrm{tan}\:\theta+\mathrm{ln}\:\mid\mathrm{sec}\:\theta+\mathrm{tan}\:\theta\mid+{c} \\ $$$$\Rightarrow\:\mathrm{2}\int_{\mathrm{0}} ^{\:\pi/\mathrm{4}} \mathrm{sec}\:^{\mathrm{3}} \theta=\sqrt{\mathrm{2}}+\mathrm{ln}\:\left(\sqrt{\mathrm{2}}+\mathrm{1}\right) \\ $$$$\:{a}^{\mathrm{2}} {r}_{{avg}} =\frac{\mathrm{2}\sqrt{\mathrm{2}}\pi{a}^{\mathrm{3}} }{\mathrm{6}}−\mathrm{2}{a}^{\mathrm{3}} \left[\frac{\pi}{\mathrm{4}}×\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\:\pi/\mathrm{4}} \mathrm{sec}\:^{\mathrm{3}} \theta{d}\theta\right] \\ $$$$\:{a}^{\mathrm{2}} {r}_{{avg}} =\frac{\mathrm{2}\sqrt{\mathrm{2}}\pi{a}^{\mathrm{3}} }{\mathrm{6}}−\mathrm{2}{a}^{\mathrm{3}} \left[\frac{\pi}{\mathrm{4}}×\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{6}}\left(\sqrt{\mathrm{2}}+\mathrm{ln}\:\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\right)\right] \\ $$$$\:\Rightarrow\:{r}_{{avg}} =\frac{{a}}{\mathrm{3}}\left[\sqrt{\mathrm{2}}+\mathrm{ln}\:\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\right] \\ $$$$\:\Rightarrow\:\:\:{r}_{{avg}} \:\approx\:\mathrm{0}.\mathrm{765}{a}\:\blacksquare \\ $$
Commented by mr W last updated on 08/Mar/19
$${thanks}\:{alot}\:{sir}! \\ $$
Answered by mr W last updated on 09/Mar/19
![for P(h,k)=O: A(x,y)=(r,θ) d=(√(x^2 +y^2 ))=r d_(Av) =((2∫_0 ^(π/4) ∫_0 ^(a/(cos θ)) r^2 drdθ)/a^2 ) d_(Av) =((2a)/3)∫_0 ^(π/4) (1/(cos^3 θ))dθ d_(Av) =((2a)/3)∫_0 ^(π/4) (1/((1−sin^2 θ)^2 ))dsin θ d_(Av) =(a/6)[ln ((1+sin θ)/(1−sin θ))+((2 tan θ)/(cos θ))]_0 ^(π/4) d_(Av) =(a/6)(ln ((2+(√2))/(2−(√2)))+2(√2)) d_(Av) =(a/3)[ln(1+(√2))+(√2)]≈0.765a](Q56026.png)
$${for}\:{P}\left({h},{k}\right)={O}: \\ $$$${A}\left({x},{y}\right)=\left({r},\theta\right) \\ $$$${d}=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }={r} \\ $$$${d}_{{Av}} =\frac{\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \int_{\mathrm{0}} ^{\frac{{a}}{\mathrm{cos}\:\theta}} {r}^{\mathrm{2}} {drd}\theta}{{a}^{\mathrm{2}} } \\ $$$${d}_{{Av}} =\frac{\mathrm{2}{a}}{\mathrm{3}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{3}} \:\theta}{d}\theta \\ $$$${d}_{{Av}} =\frac{\mathrm{2}{a}}{\mathrm{3}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\theta\right)^{\mathrm{2}} }{d}\mathrm{sin}\:\theta \\ $$$${d}_{{Av}} =\frac{{a}}{\mathrm{6}}\left[\mathrm{ln}\:\frac{\mathrm{1}+\mathrm{sin}\:\theta}{\mathrm{1}−\mathrm{sin}\:\theta}+\frac{\mathrm{2}\:\mathrm{tan}\:\theta}{\mathrm{cos}\:\theta}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$${d}_{{Av}} =\frac{{a}}{\mathrm{6}}\left(\mathrm{ln}\:\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}−\sqrt{\mathrm{2}}}+\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$$${d}_{{Av}} =\frac{{a}}{\mathrm{3}}\left[\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)+\sqrt{\mathrm{2}}\right]\approx\mathrm{0}.\mathrm{765}{a} \\ $$