let D ={(x,y)∈ R^2 /x^2 +y^2 <1} find the value of ∫∫


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Question Number 36192 by prof Abdo imad last updated on 30/May/18
$let D ={(x,y)∈ R^2 /x^2 +y^2 <1} find the value of ∫∫_D ((dxdy)/(x^2 +y^(2 ) + 2))$
$l e t D = {(x, y) \in R^{2} / x^{2} + y^{2} < 1}$ $f i n d t h e v a l u e o f \int \int_{D} \frac{d x d y}{x^{2} + y^{2} + 2}$
Commented byabdo imad last updated on 31/May/18
$let consider the diffeomorphism (r,θ)→(x,y) / x =r cosθ and y =r sinθ we have ∫∫_D ((dxdy)/(x^2 +y^2 +2)) =∫∫_(0<r<1 and 0≤θ≤2π) ((r drdθ)/(r^2 +2)) =∫_0 ^1 ((rdr)/(r^2 +2)) .∫_0 ^(2π) dθ =2π [(1/2)ln(r^2 +2)]_0 ^1 =π {ln(3)−ln(2)}.$
$l e t c o n s i d e r t h e d i f f e o m o r p h i s m (r, θ) \to (x, y) /$ $x = r c o s θ a n d y = r s i n θ w e h a v e$ $\int \int_{D} \frac{d x d y}{x^{2} + y^{2} + 2} = \int \int_{0 < r < 1 a n d 0 ⩽ θ ⩽ 2 π} \frac{r d r d θ}{r^{2} + 2}$ $= \int_{0}^{1} \frac{r d r}{r^{2} + 2} . \int_{0}^{2 π} d θ = 2 π {[\frac{1}{2} l n (r^{2} + 2)]}_{0}^{1}$ $= π {l n (3) - l n (2)} .$
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