let f(x,y)=(x+y)(√(x+y−1)) calculate ∫∫_D f(x,y)dxdy with D ={(x,y)∈R^2 / 1≤x≤2 and 1≤y≤(√3)}


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Question Number 65401 by mathmax by abdo last updated on 29/Jul/19
$let f(x,y)=(x+y)(√(x+y−1)) calculate ∫∫_D f(x,y)dxdy with D ={(x,y)∈R^2 / 1≤x≤2 and 1≤y≤(√3)}$
$l e t f (x, y) = (x + y) \sqrt{x + y - 1}$ $c a l c u l a t e \int \int_{D} f (x, y) d x d y w i t h$ $D = {(x, y) \in R^{2} / 1 ⩽ x ⩽ 2 a n d 1 ⩽ y ⩽ \sqrt{3}}$
Commented by mathmax by abdo last updated on 01/Aug/19
$∫∫_D f(x,y)dxdy =∫_1 ^(√3) (∫_1 ^2 (x+y)(√(x+y−1))dx)dy=∫_1 ^(√3) A(y)dy changement (√(x+y−1))=t give x+y−1=t^2 ⇒x =t^2 −y+1 ⇒ A(y) =∫_(√y) ^(√(y+1)) (t^2 −y+1+y)t(2t)dt =2 ∫_(√y) ^(√(y+1)) (t^4 +t^2 )dt =2[(t^5 /5)+(t^3 /3)]_(√y) ^(√(y+1)) =2{((((√(y+1)))^5 )/5)+((((√(y+1)))^3 )/3) −((((√y))^5 )/5)−((((√y))^3 )/3) =2{ (((y+1)^2 (√(y+1)))/5) +(((y+1)(√(y+1)))/3)−((y^2 (√y))/5)−((y(√y))/3)} ⇒ ∫∫_D f(x,y)dxdy =(2/5)∫_1 ^(√3) (y+1)^2 (√(y+1)) dy +(2/3)∫_1 ^(√3) (y+1)(√(y+1))dy −(2/5) ∫_1 ^(√3) y^2 (√y)dy −(2/3) ∫_1 ^(√3) y(√y)dy changement (√(y+1))=u give y+1 =u^2 ⇒ ∫_1 ^(√3) (y+1)^2 (√(y+1))dy =∫_(√2) ^(√((√3)+1)) u^4 u (2u)du =2 ∫_(√2) ^(√((√3)+1)) u^6 du =(2/7)[u^7 ]_(√2) ^(√((√3)+1)) =(2/7){ ((√((√3)+1)))^7 −((√2))^7 } ∫_1 ^(√3) (y+1)(√(y+1))dy =∫_(√2) ^(√((√3)+1)) u^2 u(2udu) =2 ∫_(√2) ^(√((√3)+1)) u^4 du =(2/5){((√((√3)+1)))^5 −((√2))^5 } also changement (√y)=u give y=u^2 ⇒∫_1 ^(√3) y^2 (√y)dy =∫_1 ^(√(√3)) u^4 (u)(2udu) =2 ∫_1 ^(√(√3)) u^6 du =(2/7)[((√(√3)))^7 −1) ∫_1 ^(√3) y(√y)dy =∫_1 ^(√(√3)) u^2 u(2u)du =2 ∫_1 ^(√(√3)) u^4 xu =(2/5){ ((√(√3)))^5 −1} the value of this integral is known$
$\int \int_{D} f (x, y) d x d y = \int_{1}^{\sqrt{3}} (\int_{1}^{2} (x + y) \sqrt{x + y - 1} d x) d y = \int_{1}^{\sqrt{3}} A (y) d y$ $c h a n g e m e n t \sqrt{x + y - 1} = t g i v e x + y - 1 = t^{2} \Rightarrow x = t^{2} - y + 1 \Rightarrow$ $A (y) = \int_{\sqrt{y}}^{\sqrt{y + 1}} (t^{2} - y + 1 + y) t (2 t) d t$ $= 2 \int_{\sqrt{y}}^{\sqrt{y + 1}} (t^{4} + t^{2}) d t = 2 {[\frac{t^{5}}{5} + \frac{t^{3}}{3}]}_{\sqrt{y}}^{\sqrt{y + 1}} = 2 {\frac{{(\sqrt{y + 1})}^{5}}{5} + \frac{{(\sqrt{y + 1})}^{3}}{3}$ $- \frac{{(\sqrt{y})}^{5}}{5} - \frac{{(\sqrt{y})}^{3}}{3} = 2 {\frac{{(y + 1)}^{2} \sqrt{y + 1}}{5} + \frac{(y + 1) \sqrt{y + 1}}{3} - \frac{y^{2} \sqrt{y}}{5} - \frac{y \sqrt{y}}{3}} \Rightarrow$ $\int \int_{D} f (x, y) d x d y = \frac{2}{5} \int_{1}^{\sqrt{3}} {(y + 1)}^{2} \sqrt{y + 1} d y + \frac{2}{3} \int_{1}^{\sqrt{3}} (y + 1) \sqrt{y + 1} d y$ $- \frac{2}{5} \int_{1}^{\sqrt{3}} y^{2} \sqrt{y} d y - \frac{2}{3} \int_{1}^{\sqrt{3}} y \sqrt{y} d y$ $c h a n g e m e n t \sqrt{y + 1} = u g i v e y + 1 = u^{2} \Rightarrow$ $\int_{1}^{\sqrt{3}} {(y + 1)}^{2} \sqrt{y + 1} d y = \int_{\sqrt{2}}^{\sqrt{\sqrt{3} + 1}} u^{4} u (2 u) d u = 2 \int_{\sqrt{2}}^{\sqrt{\sqrt{3} + 1}} u^{6} d u$ $= \frac{2}{7} {[u^{7}]}_{\sqrt{2}}^{\sqrt{\sqrt{3} + 1}} = \frac{2}{7} {{(\sqrt{\sqrt{3} + 1})}^{7} - {(\sqrt{2})}^{7}}$ $\int_{1}^{\sqrt{3}} (y + 1) \sqrt{y + 1} d y = \int_{\sqrt{2}}^{\sqrt{\sqrt{3} + 1}} u^{2} u (2 u d u) = 2 \int_{\sqrt{2}}^{\sqrt{\sqrt{3} + 1}} u^{4} d u$ $= \frac{2}{5} {{(\sqrt{\sqrt{3} + 1})}^{5} - {(\sqrt{2})}^{5}} a l s o c h a n g e m e n t \sqrt{y} = u g i v e y = u^{2}$ $\Rightarrow \int_{1}^{\sqrt{3}} y^{2} \sqrt{y} d y = \int_{1}^{\sqrt{\sqrt{3}}} u^{4} (u) (2 u d u) = 2 \int_{1}^{\sqrt{\sqrt{3}}} u^{6} d u = \frac{2}{7} [{(\sqrt{\sqrt{3}})}^{7} - 1)$ $\int_{1}^{\sqrt{3}} y \sqrt{y} d y = \int_{1}^{\sqrt{\sqrt{3}}} u^{2} u (2 u) d u = 2 \int_{1}^{\sqrt{\sqrt{3}}} u^{4} x u = \frac{2}{5} {{(\sqrt{\sqrt{3}})}^{5} - 1}$ $t h e v a l u e o f t h i s i n t e g r a l i s k n o w n$
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