Question Number 72796 by Learner-123 last updated on 03/Nov/19
$${Integrate}\:{f}\left({x},{y}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{over} \\ $$$${the}\:{triangle}\:{with}\:{vertices}\:\left(\mathrm{0},\mathrm{0}\right)\:,\left(\mathrm{1},\mathrm{0}\right), \\ $$$$\left(\mathrm{1},\sqrt{\mathrm{3}}\right)\:{after}\:{changing}\:{it}\:{to}\:{polar}\:{form}. \\ $$
Answered by mind is power last updated on 03/Nov/19
![the triagle can bee expreced As={(x,y)∈IR^2 ∣ 0≤ y≤(√3)x,0≤x≤1} x=rcos(θ), y=rsin(θ) 0≤rsin(θ)≤rcos(θ)(√3) ⇒0≤tg(θ)≤(√3)....1 0≤rcos(θ)≤1⇒0≤r≤(1/(cos(θ)))...2 1⇒θ∈[0,(π/3)] ∫∫f(x,y)dx=∫_0 ^(π/3) .∫_0 ^(1/(cos(θ))) .((rdrdθ)/((1+r^2 )^2 ))=∫_0 ^(π/3) (∫_0 ^(1/(cos(θ))) (r/((1+r^2 )^2 )) )dθ =∫_0 ^(π/3) [−(1/2).(1/(1+r^2 ))]_0 ^(1/(cos(θ))) dθ=∫_0 ^(π/3) [−((cos^2 (θ))/(2(1+cos^2 (θ))))+(1/2)]dθ ∫_0 ^(π/3) ((dθ/(2(1+cos^2 (θ)))))=∫_0 ^(π/3) (((1+tg^2 (θ)))/(4+2tg^2 (θ)))dθ=(1/(2(√2)))∫_0 ^(π/3) .(((1/( (√2)))(1+tg^2 (θ)))/((1+(((tg(θ))/( (√2))))^2 ))) =(1/(2(√2)))[arctan((√2)tg(θ))]_0 ^(π/3) =((arctan((√6)))/(2(√2)))](https://www.tinkutara.com/question/Q72802.png)
$$\:\:\:\:\mathrm{the}\:\mathrm{triagle}\:\mathrm{can}\:\mathrm{bee}\:\mathrm{expreced}\:\mathrm{As}=\left\{\left(\mathrm{x},\mathrm{y}\right)\in\mathrm{IR}^{\mathrm{2}} \mid\:\:\:\mathrm{0}\leqslant\:\mathrm{y}\leqslant\sqrt{\mathrm{3}}\mathrm{x},\mathrm{0}\leqslant\mathrm{x}\leqslant\mathrm{1}\right\} \\ $$$$\:\mathrm{x}=\mathrm{rcos}\left(\theta\right), \\ $$$$\mathrm{y}=\mathrm{rsin}\left(\theta\right)\: \\ $$$$\mathrm{0}\leqslant\mathrm{rsin}\left(\theta\right)\leqslant\mathrm{rcos}\left(\theta\right)\sqrt{\mathrm{3}}\:\Rightarrow\mathrm{0}\leqslant\mathrm{tg}\left(\theta\right)\leqslant\sqrt{\mathrm{3}}….\mathrm{1} \\ $$$$\mathrm{0}\leqslant\mathrm{rcos}\left(\theta\right)\leqslant\mathrm{1}\Rightarrow\mathrm{0}\leqslant\mathrm{r}\leqslant\frac{\mathrm{1}}{\mathrm{cos}\left(\theta\right)}…\mathrm{2} \\ $$$$\mathrm{1}\Rightarrow\theta\in\left[\mathrm{0},\frac{\pi}{\mathrm{3}}\right] \\ $$$$\int\int\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\mathrm{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} .\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{cos}\left(\theta\right)}} .\frac{\mathrm{rdrd}\theta}{\left(\mathrm{1}+\mathrm{r}^{\mathrm{2}} \right)^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \left(\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{cos}\left(\theta\right)}} \frac{\mathrm{r}}{\left(\mathrm{1}+\mathrm{r}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\right)\mathrm{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \left[−\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{1}+\mathrm{r}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{cos}\left(\theta\right)}} \mathrm{d}\theta=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \left[−\frac{\mathrm{cos}^{\mathrm{2}} \left(\theta\right)}{\mathrm{2}\left(\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \left(\theta\right)\right)}+\frac{\mathrm{1}}{\mathrm{2}}\right]\mathrm{d}\theta \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \left(\frac{\mathrm{d}\theta}{\mathrm{2}\left(\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \left(\theta\right)\right)}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \frac{\left(\mathrm{1}+\mathrm{tg}^{\mathrm{2}} \left(\theta\right)\right)}{\mathrm{4}+\mathrm{2tg}^{\mathrm{2}} \left(\theta\right)}\mathrm{d}\theta=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} .\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\mathrm{1}+\mathrm{tg}^{\mathrm{2}} \left(\theta\right)\right)}{\left(\mathrm{1}+\left(\frac{\mathrm{tg}\left(\theta\right)}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left[\mathrm{arctan}\left(\sqrt{\mathrm{2}}\mathrm{tg}\left(\theta\right)\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} =\frac{\mathrm{arctan}\left(\sqrt{\mathrm{6}}\right)}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Learner-123 last updated on 03/Nov/19
$${thanks}\:{sir}. \\ $$
Commented by mind is power last updated on 03/Nov/19
$$\mathrm{y}'\mathrm{re}\:\mathrm{welcom} \\ $$