Menu Close

1-calculate-A-n-1-n-2-sin-x-2-3y-2-e-x-2-3y-2-dxdy-2-determine-lim-n-A-n-




Question Number 65398 by mathmax by abdo last updated on 29/Jul/19
1) calculate  A_n =∫∫_([1,n[^2 )      sin(x^2  +3y^2 ) e^(−x^2 −3y^2 ) dxdy  2) determine lim_(n→+∞)  A_n
1)calculateAn=[1,n[2sin(x2+3y2)ex23y2dxdy2)determinelimn+An
Commented by mathmax by abdo last updated on 31/Jul/19
1) let consider thediffeomorphism (r,θ)→(x,y)=(rcosθ,(r/( (√3)))sinθ)  ϕ(r,θ) =(ϕ_1 ,ϕ_2 ) / ϕ_1 (r,θ)=rcosθ and ϕ_2 (r,θ)=(r/( (√3)))sinθ  M_j (ϕ) = ((((∂ϕ_1 /∂r)         (∂ϕ_1 /∂θ))),(((∂ϕ_2 /∂r)            (∂ϕ_2 /∂θ))) )  = (((cosθ            −rsinθ)),((((sinθ)/( (√3)))                 (r/( (√3)))cosθ)) )  and detM_j (ϕ) =(r/( (√3)))  we have   1≤x<n  and  1≤y<n ⇒  1≤x^2 <n^2  and3≤3y^2 <3n^2  ⇒4≤x^2  +3y^2 <4n^(2 )  ⇒2≤r<2n  A_n =∫∫_(2≤r<2n and 0≤θ≤(π/2))     sin(r^2 )e^(−r^2 ) (r/( (√3))) drdθ  =(π/2) ∫_2 ^(2n)  r sin(r^2 )e^(−r^2 ) dr  by parts u^′  =r sin(r^2 ) and v =e^(−r^2 )  ⇒  ∫_2 ^(2n)   rsin(r^2 )e^(−r^2 ) dr =[−(1/2)cos(r^2 )e^(−r^2 ) ]_2 ^(2n)  −∫_2 ^(2n) (−(1/2)cos(r^2 ))(−2r)e^(−r^2 ) dr  =−(1/2){cos(4n^2 )e^(−4n^2 )  −cos(4)e^(−4) }−∫_2 ^(2n)  r cos(r^2 ) e^(−r^2 ) dr  again by parts  u^′  =rcos(r^2 ) and v =e^(−r^2 ) ⇒  ∫_2 ^(2n)  rcos(r^2 )e^(−r^2 ) dr =[(1/2)sin(r^2 )e^(−r^2 ) ]_2 ^(2n)  −∫_2 ^(2n) (1/2)sin(r^2 )(−2r)e^(−r^2 ) dr  =(1/2){sin(4n^2 )e^(−4n^2 ) −sin4 e^(−4) } +∫_2 ^(2n)  rsin(r^2 )e^(−r^2 ) dr ⇒  2∫_2 ^(2n)  rsin(r^2 )e^(−r^2 ) dr =(1/2){cos(4)e^(−4) −cos(4n^2 )e^(−4n^2 ) }  +(1/2){sin(4)e^(−4) −sin(4n^2 )e^(−4n^2 ) } ⇒  A_n =(π/8){  e^(−4) (cos(4)+sin4)−e^(−4n^2 ) (cos(4n^2 )+sin(4n^2 )}  2) lim_(n→+∞)   A_n =((πe^(−4) )/8){ cos(4)+sin(4)}
1)letconsiderthediffeomorphism(r,θ)(x,y)=(rcosθ,r3sinθ)φ(r,θ)=(φ1,φ2)/φ1(r,θ)=rcosθandφ2(r,θ)=r3sinθMj(φ)=(φ1rφ1θφ2rφ2θ)=(cosθrsinθsinθ3r3cosθ)anddetMj(φ)=r3wehave1x<nand1y<n1x2<n2and33y2<3n24x2+3y2<4n22r<2nAn=2r<2nand0θπ2sin(r2)er2r3drdθ=π222nrsin(r2)er2drbypartsu=rsin(r2)andv=er222nrsin(r2)er2dr=[12cos(r2)er2]22n22n(12cos(r2))(2r)er2dr=12{cos(4n2)e4n2cos(4)e4}22nrcos(r2)er2dragainbypartsu=rcos(r2)andv=er222nrcos(r2)er2dr=[12sin(r2)er2]22n22n12sin(r2)(2r)er2dr=12{sin(4n2)e4n2sin4e4}+22nrsin(r2)er2dr222nrsin(r2)er2dr=12{cos(4)e4cos(4n2)e4n2}+12{sin(4)e4sin(4n2)e4n2}An=π8{e4(cos(4)+sin4)e4n2(cos(4n2)+sin(4n2)}2)limn+An=πe48{cos(4)+sin(4)}

Leave a Reply

Your email address will not be published. Required fields are marked *