calculate ∫∫_([0,1]^2 ) (√(x^2 +y^2 ))sin((√(x^2 +y^2 )))dxdy


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Question Number 62197 by maxmathsup by imad last updated on 17/Jun/19

$c a l c u l a t e \int \int_{{[0, 1]}^{2}} \sqrt{x^{2} + y^{2}} s i n (\sqrt{x^{2} + y^{2}}) d x d y$
Commented by mathmax by abdo last updated on 21/Jun/19
$let use the diffeomorphism x =rcosθ and y =rsinθ 0≤x≤1 and 0≤y≤1 ⇒0≤x^2 +y^2 ≤2 ⇒0≤r^2 ≤2 ⇒0≤r≤(√2) A=∫∫_([0,1]^2 ) (√(x^2 +y^2 ))sin((√(x^2 +y^2 )))dxdy =∫∫_(0≤r≤(√2) and 0≤θ≤(π/2)) r sin(r)rdrdθ =∫_0 ^(√2) r^2 sin(r)dr ∫_0 ^(π/2) dθ =(π/2) ∫_0 ^(√2) r^2 sin(r)dr by parts u=r^2 and v^′ =sinr ∫_0 ^(√2) r^2 sin(r)dr =[−r^2 cos(r)]_0 ^(√2) +∫_0 ^(√2) 2r cosrdr =−2cos((√2)) +2 { [r sin(r)]_0 ^(√2) −∫_0 ^(√2) sinr dr} =−2cos((√2)) +2{ (√2)sin((√2)) +[cosr]_0 ^(√2) } =−2cos((√2)) +2(√2)sin((√2)) +2(cos((√2))−1) ⇒ A =(π/2){ −2cos((√2))+2(√2)sin((√2)) +2 cos((√2))−2}=π{(√2)sin((√2))−1} .$
$l e t u s e t h e d i f f e o m o r p h i s m x = r c o s θ a n d y = r s i n θ$ $0 ⩽ x ⩽ 1 a n d 0 ⩽ y ⩽ 1 \Rightarrow 0 ⩽ x^{2} + y^{2} ⩽ 2 \Rightarrow 0 ⩽ r^{2} ⩽ 2 \Rightarrow 0 ⩽ r ⩽ \sqrt{2}$ $A = \int \int_{{[0, 1]}^{2}} \sqrt{x^{2} + y^{2}} s i n (\sqrt{x^{2} + y^{2}}) d x d y = \int \int_{0 ⩽ r ⩽ \sqrt{2} a n d 0 ⩽ θ ⩽ \frac{π}{2}} r s i n (r) r d r d θ$ $= \int_{0}^{\sqrt{2}} r^{2} s i n (r) d r \int_{0}^{\frac{π}{2}} d θ = \frac{π}{2} \int_{0}^{\sqrt{2}} r^{2} s i n (r) d r b y p a r t s u = r^{2} a n d v^{^{'}} = s i n r$ $\int_{0}^{\sqrt{2}} r^{2} s i n (r) d r = {[- r^{2} c o s (r)]}_{0}^{\sqrt{2}} + \int_{0}^{\sqrt{2}} 2 r c o s r d r$ $= - 2 c o s (\sqrt{2}) + 2 {{[r s i n (r)]}_{0}^{\sqrt{2}} - \int_{0}^{\sqrt{2}} s i n r d r}$ $= - 2 c o s (\sqrt{2}) + 2 {\sqrt{2} s i n (\sqrt{2}) + {[c o s r]}_{0}^{\sqrt{2}}}$ $= - 2 c o s (\sqrt{2}) + 2 \sqrt{2} s i n (\sqrt{2}) + 2 (c o s (\sqrt{2}) - 1) \Rightarrow$ $A = \frac{π}{2} {- 2 c o s (\sqrt{2}) + 2 \sqrt{2} s i n (\sqrt{2}) + 2 c o s (\sqrt{2}) - 2} = π {\sqrt{2} s i n (\sqrt{2}) - 1} .$
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