calculate ∫∫_D xy(√(x^2 +2y^2 ))dxdy D={(x,y)/0≤x≤1 and 0≤y≤(√(1−x^2 ))}


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Question Number 78284 by msup trace by abdo last updated on 15/Jan/20
$calculate ∫∫_D xy(√(x^2 +2y^2 ))dxdy D={(x,y)/0≤x≤1 and 0≤y≤(√(1−x^2 ))}$
$c a l c u l a t e \int \int_{D} x y \sqrt{x^{2} + 2 y^{2}} d x d y$ $D = {(x, y) / 0 ⩽ x ⩽ 1 a n d 0 ⩽ y ⩽ \sqrt{1 - x^{2}}}$
Commented by abdomathmax last updated on 17/Jan/20
$I =∫∫_D xy(√(x^2 +2y^2 ))dxdy ⇒ I =∫_0 ^1 (∫_0 ^(√(1−x^2 )) y(√(x^2 +2y^2 ))dy)xdx but ∫_0 ^(√(1−x^2 )) y(√(x^2 +2y^2 ))dy =[(1/6)(x^2 +2y^2 )^(3/2) ]_0 ^(√(1−x^2 )) =(1/6){ (x^2 +(x^2 +2−2x^2 )^(3/2) −x^3 } =(1/6){ (x^2 −x^3 +(2−x^2 )^(3/2) } ⇒ 6I =∫_0 ^1 x(x^2 −x^3 +(2−x^2 )^(3/2) )dx =∫_0 ^1 (x^3 −x^4 )dx+∫_0 ^1 x(2−x^2 )^(3/2) dx =[(x^4 /4)−(x^5 /5)]_0 ^1 +∫_0 ^1 x(2−x^2 )^(3/2) dx =(1/(20)) +∫_0 ^1 x(2−x^2 )^(3/2) dx we have ∫_0 ^1 x(2−x^2 )^(3/2) dx =_(x=(√2)sint) ∫_0 ^(π/4) (√2)sint(2−2sin^2 t)^(3/2) (√2)costdt =2×2^(3/2) ∫_0 ^(π/4) sint cos^4 t dt =2^(5/2) [−(1/5)cos^5 t]_0 ^(π/4) =−((4(√2))/5)( ((1/(√2)))^5 −1) =−((4(√2))/5)( (1/(4(√2)))−1) =−(1/5)+((4(√2))/5) ⇒ 6I =(1/(20))−(1/5) +((4(√2))/5) =((−3)/(20)) +((4(√2))/5) ⇒ I =−(1/(40)) +((2(√2))/(15))$
$I = \int \int_{D} x y \sqrt{x^{2} + 2 y^{2}} d x d y \Rightarrow$ $I = \int_{0}^{1} (\int_{0}^{\sqrt{1 - x^{2}}} y \sqrt{x^{2} + 2 y^{2}} d y) x d x b u t$ $\int_{0}^{\sqrt{1 - x^{2}}} y \sqrt{x^{2} + 2 y^{2}} d y = {[\frac{1}{6} {(x^{2} + 2 y^{2})}^{\frac{3}{2}}]}_{0}^{\sqrt{1 - x^{2}}}$ $= \frac{1}{6} {(x^{2} + {(x^{2} + 2 - 2 x^{2})}^{\frac{3}{2}} - x^{3}}$ $= \frac{1}{6} {(x^{2} - x^{3} + {(2 - x^{2})}^{\frac{3}{2}}} \Rightarrow$ $6 I = \int_{0}^{1} x (x^{2} - x^{3} + {(2 - x^{2})}^{\frac{3}{2}}) d x$ $= \int_{0}^{1} (x^{3} - x^{4}) d x + \int_{0}^{1} x {(2 - x^{2})}^{\frac{3}{2}} d x$ $= {[\frac{x^{4}}{4} - \frac{x^{5}}{5}]}_{0}^{1} + \int_{0}^{1} x {(2 - x^{2})}^{\frac{3}{2}} d x$ $= \frac{1}{20} + \int_{0}^{1} x {(2 - x^{2})}^{\frac{3}{2}} d x w e h a v e$ $\int_{0}^{1} x {(2 - x^{2})}^{\frac{3}{2}} d x =_{x = \sqrt{2} s i n t} \int_{0}^{\frac{π}{4}} \sqrt{2} s i n t {(2 - 2 {s i n}^{2} t)}^{\frac{3}{2}} \sqrt{2} c o s t d t$ $= 2 \times 2^{\frac{3}{2}} \int_{0}^{\frac{π}{4}} s i n t {c o s}^{4} t d t$ $= 2^{\frac{5}{2}} {[- \frac{1}{5} {c o s}^{5} t]}_{0}^{\frac{π}{4}} = - \frac{4 \sqrt{2}}{5} ({(\frac{1}{\sqrt{2}})}^{5} - 1)$ $= - \frac{4 \sqrt{2}}{5} (\frac{1}{4 \sqrt{2}} - 1) = - \frac{1}{5} + \frac{4 \sqrt{2}}{5} \Rightarrow$ $6 I = \frac{1}{20} - \frac{1}{5} + \frac{4 \sqrt{2}}{5} = \frac{- 3}{20} + \frac{4 \sqrt{2}}{5} \Rightarrow$ $I = - \frac{1}{40} + \frac{2 \sqrt{2}}{15}$
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