find I_n =∫∫_([1,n]^2 ) (√(x^2 +y^2 ))ln(x^2 +y^2 )dxdy


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Question Number 78276 by msup trace by abdo last updated on 15/Jan/20

$f i n d I_{n} = \int \int_{{[1, n]}^{2}} \sqrt{x^{2} + y^{2}} l n (x^{2} + y^{2}) d x d y$
Commented by mathmax by abdo last updated on 17/Jan/20
$let consider the diffeomorphism (r,θ)→(x,y)=(rcosθ,rsinθ) we have 1≤x≤n and 1≤y≤n ⇒2≤x^2 +y^2 ≤2n^2 ⇒(√2)≤r≤n(√2) and 0≤θ≤(π/2) ⇒ I_n =∫_0 ^(π/2) ∫_(√2) ^(n(√2)) rln(r^2 )rdr dθ =π ∫_(√2) ^(n(√2)) r^2 ln(r)dr and by parts ∫_(√2) ^(n(√2)) r^2 ln(r)dr =[(r^3 /3)ln(r)]_(√2) ^(n(√2)) −∫_(√2) ^(n(√2)) (r^3 /3)(dr/r) =(1/3)(2(√2)n^3 ln(n(√2))−((2(√2))/3)ln((√2))) −(1/3)∫_(√2) ^(n(√2)) r^2 dr =((2(√2))/3){ n^3 ln(n(√2))−ln((√2))}−(1/9)[r^3 ]_(√2) ^(n(√2)) ⇒ I_n =((2(√2))/3){n^3 ln(n(√2))−ln((√2))}−(1/9){2(√2)n^3 −2(√2)}$
$l e t c o n s i d e r t h e d i f f e o m o r p h i s m (r, θ) \to (x, y) = (r c o s θ, r s i n θ)$ $w e h a v e 1 ⩽ x ⩽ n a n d 1 ⩽ y ⩽ n \Rightarrow 2 ⩽ x^{2} + y^{2} ⩽ 2 n^{2} \Rightarrow \sqrt{2} ⩽ r ⩽ n \sqrt{2}$ $a n d 0 ⩽ θ ⩽ \frac{π}{2} \Rightarrow I_{n} = \int_{0}^{\frac{π}{2}} \int_{\sqrt{2}}^{n \sqrt{2}} r l n (r^{2}) r d r d θ$ $= π \int_{\sqrt{2}}^{n \sqrt{2}} r^{2} l n (r) d r a n d b y p a r t s \int_{\sqrt{2}}^{n \sqrt{2}} r^{2} l n (r) d r$ $= {[\frac{r^{3}}{3} l n (r)]}_{\sqrt{2}}^{n \sqrt{2}} - \int_{\sqrt{2}}^{n \sqrt{2}} \frac{r^{3}}{3} \frac{d r}{r} = \frac{1}{3} (2 \sqrt{2} n^{3} l n (n \sqrt{2}) - \frac{2 \sqrt{2}}{3} l n (\sqrt{2}))$ $- \frac{1}{3} \int_{\sqrt{2}}^{n \sqrt{2}} r^{2} d r = \frac{2 \sqrt{2}}{3} {n^{3} l n (n \sqrt{2}) - l n (\sqrt{2})} - \frac{1}{9} {[r^{3}]}_{\sqrt{2}}^{n \sqrt{2}} \Rightarrow$ $I_{n} = \frac{2 \sqrt{2}}{3} {n^{3} l n (n \sqrt{2}) - l n (\sqrt{2})} - \frac{1}{9} {2 \sqrt{2} n^{3} - 2 \sqrt{2}}$
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