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-

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) \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} \dots . 1

0 ⩽ rcos (θ) ⩽ 1 \Rightarrow 0 ⩽ r ⩽ \frac{1}{\cos (θ)} \dots 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