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


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Question Number 61648 by maxmathsup by imad last updated on 05/Jun/19
$calculate ∫∫_W (x^2 −2y^2 )(√(x^2 +y^2 +3))dxdy with W ={ (x,y) ∈ R^2 / 1≤x ≤(√3) and x^2 +y^2 −2y ≤ 2 }$
$c a l c u l a t e \int \int_{W} (x^{2} - 2 y^{2}) \sqrt{x^{2} + y^{2} + 3} d x d y w i t h$ $W = {(x, y) \in R^{2} / 1 ⩽ x ⩽ \sqrt{3} a n d x^{2} + y^{2} - 2 y ⩽ 2}$
Commented by maxmathsup by imad last updated on 06/Jun/19
$we have x^2 +y^2 −2y ≤2 ⇒x^2 +y^2 −2y +1−1 ≤2 ⇒x^2 +(y−1)^2 ≤3 let use the diffeomorphism x=rcosθ and y−1 =rsinθ we have 1≤x^2 ≤3 ⇒1≤x^2 +(y−1)^2 ≤6 ⇒1≤r^2 ≤6 ⇒1≤r≤(√6) ∫∫_W (x^2 −2y^2 )(√(x^2 +y^2 +3))dxdy =∫∫_(1≤r≤(√6) and −(π/2)≤θ≤(π/2)) (r^2 cos^2 θ−2r^2 sin^2 θ)(√(r^2 +3))rdrdθ =∫_1 ^(√6) r^3 (√(3+r^2 ))dr ∫_(−(π/2)) ^(π/2) (cos^2 θ −2sin^2 θ)dθ ∫_(−(π/2)) ^(π/2) (cos^2 θ −2sin^2 θ)dθ =∫_(−(π/2)) ^(π/2) (((1+cos(2θ))/2) −(1−cos(2θ))dθ =(1/2) ∫_(−((.π)/2)) ^(π/2) (1+cos(2θ)−2 +2cos(2θ))dθ =(1/4) ∫_0 ^(π/2) {3cosθ −1}dθ = (3/4)[sinθ]_0 ^(π/2) −(π/8) =(3/4) −(π/8) . chang.(√(3+r^2 ))=t give 3+r^2 =t^2 ⇒rdr =tdt ∫_1 ^(√6) r^3 (√(3+r^2 ))dr = ∫_2 ^3 (t^2 −3)t dt = ∫_2 ^3 (t^3 −3t)dt =[(1/4)t^4 −(3/2)t^2 ]_2 ^3 =(3^4 /4) −(3^3 /2) −(2^4 /2) +(3/2).2^2 =((81)/4) −((27)/2) −8 +6 =((81−54)/4) −2 =((81−62)/4) =((18)/4) =(9/2) ⇒ ⇒ I = (9/2)((3/4)−(π/8)) =((27)/8) −((9π)/(16)) .$
$w e h a v e x^{2} + y^{2} - 2 y ⩽ 2 \Rightarrow x^{2} + y^{2} - 2 y + 1 - 1 ⩽ 2 \Rightarrow x^{2} + {(y - 1)}^{2} ⩽ 3$ $l e t u s e t h e d i f f e o m o r p h i s m x = r c o s θ a n d y - 1 = r s i n θ$ $w e h a v e 1 ⩽ x^{2} ⩽ 3 \Rightarrow 1 ⩽ x^{2} + {(y - 1)}^{2} ⩽ 6 \Rightarrow 1 ⩽ r^{2} ⩽ 6 \Rightarrow 1 ⩽ r ⩽ \sqrt{6}$ $\int \int_{W} (x^{2} - 2 y^{2}) \sqrt{x^{2} + y^{2} + 3} d x d y = \int \int_{1 ⩽ r ⩽ \sqrt{6} a n d - \frac{π}{2} ⩽ θ ⩽ \frac{π}{2}} (r^{2} {c o s}^{2} θ - 2 r^{2} {s i n}^{2} θ) \sqrt{r^{2} + 3} r d r d θ$ $= \int_{1}^{\sqrt{6}} r^{3} \sqrt{3 + r^{2}} d r \int_{- \frac{π}{2}}^{\frac{π}{2}} ({c o s}^{2} θ - 2 {s i n}^{2} θ) d θ$ $\int_{- \frac{π}{2}}^{\frac{π}{2}} ({c o s}^{2} θ - 2 {s i n}^{2} θ) d θ = \int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{1 + c o s (2 θ)}{2} - (1 - c o s (2 θ)) d θ$ $= \frac{1}{2} \int_{- \frac{. π}{2}}^{\frac{π}{2}} (1 + c o s (2 θ) - 2 + 2 c o s (2 θ)) d θ = \frac{1}{4} \int_{0}^{\frac{π}{2}} {3 c o s θ - 1} d θ$ $= \frac{3}{4} {[s i n θ]}_{0}^{\frac{π}{2}} - \frac{π}{8} = \frac{3}{4} - \frac{π}{8} . c h a n g . \sqrt{3 + r^{2}} = t g i v e 3 + r^{2} = t^{2} \Rightarrow r d r = t d t$ $\int_{1}^{\sqrt{6}} r^{3} \sqrt{3 + r^{2}} d r = \int_{2}^{3} (t^{2} - 3) t d t = \int_{2}^{3} (t^{3} - 3 t) d t = {[\frac{1}{4} t^{4} - \frac{3}{2} t^{2}]}_{2}^{3}$ $= \frac{3^{4}}{4} - \frac{3^{3}}{2} - \frac{2^{4}}{2} + \frac{3}{2} . 2^{2} = \frac{81}{4} - \frac{27}{2} - 8 + 6 = \frac{81 - 54}{4} - 2 = \frac{81 - 62}{4} = \frac{18}{4} = \frac{9}{2} \Rightarrow$ $\Rightarrow I = \frac{9}{2} (\frac{3}{4} - \frac{π}{8}) = \frac{27}{8} - \frac{9 π}{16} .$
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