Integrate f(x,y)=(1/((1+x^2 +y^2 )^2 )) over the triangle with vertices (0,0) ,(1,0), (1,(√3)) after changing it to polar form.


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Question Number 72796 by Learner-123 last updated on 03/Nov/19

$I n t e g r a t e f (x, y) = \frac{1}{{(1 + x^{2} + y^{2})}^{2}} o v e r$ $t h e t r i a n g l e w i t h v e r t i c e s (0, 0), (1, 0),$ $(1, \sqrt{3}) a f t e r c h a n g i n g i t t o p o l a r f o r m .$
	Answered by mind is power last updated on 03/Nov/19
	$the triagle can bee expreced As={(x,y)∈IR^2 ∣ 0≤ y≤(√3)x,0≤x≤1} x=rcos(θ), y=rsin(θ) 0≤rsin(θ)≤rcos(θ)(√3) ⇒0≤tg(θ)≤(√3)....1 0≤rcos(θ)≤1⇒0≤r≤(1/(cos(θ)))...2 1⇒θ∈[0,(π/3)] ∫∫f(x,y)dx=∫_0 ^(π/3) .∫_0 ^(1/(cos(θ))) .((rdrdθ)/((1+r^2 )^2 ))=∫_0 ^(π/3) (∫_0 ^(1/(cos(θ))) (r/((1+r^2 )^2 )) )dθ =∫_0 ^(π/3) [−(1/2).(1/(1+r^2 ))]_0 ^(1/(cos(θ))) dθ=∫_0 ^(π/3) [−((cos^2 (θ))/(2(1+cos^2 (θ))))+(1/2)]dθ ∫_0 ^(π/3) ((dθ/(2(1+cos^2 (θ)))))=∫_0 ^(π/3) (((1+tg^2 (θ)))/(4+2tg^2 (θ)))dθ=(1/(2(√2)))∫_0 ^(π/3) .(((1/(√2))(1+tg^2 (θ)))/((1+(((tg(θ))/(√2)))^2 ))) =(1/(2(√2)))[arctan((√2)tg(θ))]_0 ^(π/3) =((arctan((√6)))/(2(√2)))$
	$the triagle can bee expreced As = {(x, y) \in {IR}^{2} ∣ 0 ⩽ y ⩽ \sqrt{3} x, 0 ⩽ x ⩽ 1}$ $x = rcos (θ),$ $y = rsin (θ)$ $0 ⩽ rsin (θ) ⩽ rcos (θ) \sqrt{3} \Rightarrow 0 ⩽ tg (θ) ⩽ \sqrt{3} . . . . 1$ $0 ⩽ rcos (θ) ⩽ 1 \Rightarrow 0 ⩽ r ⩽ \frac{1}{\cos (θ)} . . . 2$ $1 \Rightarrow θ \in [0, \frac{π}{3}]$ $\int \int f (x, y) dx = \int_{0}^{\frac{π}{3}} . \int_{0}^{\frac{1}{\cos (θ)}} . \frac{rdrd θ}{{(1 + r^{2})}^{2}} = \int_{0}^{\frac{π}{3}} (\int_{0}^{\frac{1}{\cos (θ)}} \frac{r}{{(1 + r^{2})}^{2}}) d θ$ $= \int_{0}^{\frac{π}{3}} {[- \frac{1}{2} . \frac{1}{1 + r^{2}}]}_{0}^{\frac{1}{\cos (θ)}} d θ = \int_{0}^{\frac{π}{3}} [- \frac{\cos^{2} (θ)}{2 (1 + \cos^{2} (θ))} + \frac{1}{2}] d θ$ $\int_{0}^{\frac{π}{3}} (\frac{d θ}{2 (1 + \cos^{2} (θ))}) = \int_{0}^{\frac{π}{3}} \frac{(1 + {tg}^{2} (θ))}{4 + {2 tg}^{2} (θ)} d θ = \frac{1}{2 \sqrt{2}} \int_{0}^{\frac{π}{3}} . \frac{\frac{1}{\sqrt{2}} (1 + {tg}^{2} (θ))}{(1 + {(\frac{tg (θ)}{\sqrt{2}})}^{2})}$ $= \frac{1}{2 \sqrt{2}} {[\arctan (\sqrt{2} tg (θ))]}_{0}^{\frac{π}{3}} = \frac{\arctan (\sqrt{6})}{2 \sqrt{2}}$
	Commented by Learner-123 last updated on 03/Nov/19

	$t h a n k s s i r .$
	Commented by mind is power last updated on 03/Nov/19

	$y^{'} re welcom$
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