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Question Number 94992 by Mr.D.N. last updated on 22/May/20

 1. Show  that:     ∫_0 ^(  π)    ((xdx)/((a^2 sin^2 x+b^2 cos^2 x)^2 )) = ((π^2 (a^2 +b^2 ))/(4a^3 b^3 ))       2.The density at the point (x,y) of a lamina bounded by the circle        x^2 +y^2 −2ax=0 is ϱ =x   find its mass.   3.∗   4. If  z= ((cos y)/x) and x=u^2 −v , y=e^x  find (dz/dv).

1.Showthat:0πxdx(a2sin2x+b2cos2x)2=π2(a2+b2)4a3b32.Thedensityatthepoint(x,y)ofalaminaboundedbythecirclex2+y22ax=0isϱ=xfinditsmass.3.4.Ifz=cosyxandx=u2v,y=exfinddzdv.

Answered by bobhans last updated on 22/May/20

(4)z = ((cos (e^x ))/x).⇒ (dz/dx) = ((−xe^x sin (e^x )−cos (e^x ))/x^2 )  v=u^2 −x⇒(dv/dx)= −1  (dz/dv) = (dz/dx) × (dx/dv) = ((xe^x sin (e^x )+cos (e^x ))/x^2 ) ?

(4)z=cos(ex)x.dzdx=xexsin(ex)cos(ex)x2v=u2xdvdx=1dzdv=dzdx×dxdv=xexsin(ex)+cos(ex)x2?

Commented by Mr.D.N. last updated on 22/May/20

thank you nice solution��

Answered by Ar Brandon last updated on 22/May/20

1\I=∫_0 ^π ((xdx)/((a^2 sin^2 x+b^2 cos^2 x)^2 ))  x=π−u⇒dx=−du⇒I=∫_0 ^π (π/((a^2 sin^2 u+b^2 cos^2 u)^2 ))du−I  ⇒2I=∫_0 ^π (π/((a^2 sin^2 x+b^2 cos^2 x)^2 ))dx=π∫_0 ^π ((sec^4 x)/((a^2 tan^2 x+b^2 )^2 ))dx  ⇒2I=_(t=tanx) π∫_(+0) ^(−0) ((t^2 +1)/((a^2 t^2 +b^2 )^2 ))dt=π∫_(+0) ^(−0) [((1/a^2 )/(a^2 t^2 +b^2 ))+(((a^2 −b^2 )/a^2 )/((a^2 t^2 +b^2 )^2 ))]dt  2I=(π/(a^2 b^2 ))∫_(+0) ^(−0) (1/(((a/b)t)^2 +1))dt+((π(a^2 −b^2 ))/(a^2 b^4 ))∫_(+0) ^(−0) (1/([((a/b)t)^2 +1]^2 ))dt  2I=(π/(a^3 b))[tan^(−1) ((a/b)t)]_(+0) ^(−0) +((π(a^2 −b^2 ))/(a^2 b^4 ))∫_0 ^π (((b/a)sec^2 θ)/(sec^4 θ))dθ  =(π^2 /(a^3 b))+((π(a^2 −b^2 ))/(2a^3 b^3 ))∫_0 ^π (1+cos 2θ)dθ=(π^2 /(a^3 b))+((π(a^2 −b^2 ))/(2a^3 b^3 ))[θ+((sin2θ)/2)]_0 ^π   =(π^2 /(a^3 b))+((π^2 (a^2 −b^2 ))/(2a^3 b^3 ))=((2π^2 b^2 +a^2 π^2 −π^2 b^2 )/(2a^3 b^3 ))=((π^2 (a^2 +b^2 ))/(2a^3 b^3 ))  ⇒∫_0 ^π ((xdx)/((a^2 sin^2 x+b^2 cos^2 x)^2 ))=((π^2 (a^2 +b^2 ))/(4a^3 b^3 ))

1I=0πxdx(a2sin2x+b2cos2x)2x=πudx=duI=0ππ(a2sin2u+b2cos2u)2duI2I=0ππ(a2sin2x+b2cos2x)2dx=π0πsec4x(a2tan2x+b2)2dx2I=t=tanxπ+00t2+1(a2t2+b2)2dt=π+00[1/a2a2t2+b2+(a2b2)/a2(a2t2+b2)2]dt2I=πa2b2+001(abt)2+1dt+π(a2b2)a2b4+001[(abt)2+1]2dt2I=πa3b[tan1(abt)]+00+π(a2b2)a2b40π(b/a)sec2θsec4θdθ=π2a3b+π(a2b2)2a3b30π(1+cos2θ)dθ=π2a3b+π(a2b2)2a3b3[θ+sin2θ2]0π=π2a3b+π2(a2b2)2a3b3=2π2b2+a2π2π2b22a3b3=π2(a2+b2)2a3b30πxdx(a2sin2x+b2cos2x)2=π2(a2+b2)4a3b3

Commented by Mr.D.N. last updated on 22/May/20

 lim is all[right you might be mistek  some way]

limisall[rightyoumightbemisteksomeway]

Commented by Ar Brandon last updated on 22/May/20

Okay let's check my solution and see if I might have made some mistakes.

Commented by niroj last updated on 22/May/20

let me solve then you find your correction just taken few time then you find easily where is a error....��

Commented by Ar Brandon last updated on 22/May/20

Okay, I'm grateful. Thanks

Commented by Ar Brandon last updated on 22/May/20

Hey hold on. I wanna give it a last try.����

Commented by Ar Brandon last updated on 22/May/20

Thanks for your patience Mr. Niroj ��

Commented by Ar Brandon last updated on 22/May/20

I found my mistake.�� It was necessary for me to respect the signs. That is, distinguish between –0 and +0��

Answered by niroj last updated on 22/May/20

1.  I= ∫_0 ^(  π)  (( xdx)/((a^2 sin^2 x+b^2 cos^2 x)^2 )) .....(i)      = ∫_0 ^( π)  (((π−x)dx)/({a^2 sin^2 (π−a)+b^2 cos^2  (π−a)}^2 ))          I = ∫_0 ^( π)  (((π−x)dx)/((a^2 sin^2 x+b^2 cos^2 x)^2 ))....(ii)   Added (i)+(ii)    2I= ∫_0 ^π  (π/((a^2 sin^2 x+b^2 cos^2 x)^2 ))dx       I= (π/2)×2∫_0 ^( (π/2)) (( 1)/((a^2 sin^2 x+b^2 cos^2 x)^2 ))dx      = π ∫_0 ^( (π/2))  ((  1)/((a^2 sin^2 x+b^2 cos^2 x)^2 ))dx    Divide numerator& denominator by (1/(cos^4 x))      = π∫_0 ^( (π/2))  (( sec^4 xdx)/({(1/(cos^2 x))(a^2 sin^2 x+b^2 cos^2 x)}^2 ))    = π ∫_0 ^( (π/2))   (((tan^2 x+1) sec^2 x dx)/((a^2 tan^2 x+b^2 )^2 ))    Put  tan x = t         sec^2 xdx=dt    If x=(π/2) ⇒ t=∞   If x=0 ⇒  t=0     Now,     π∫_0 ^( ∞)   (((t^2 +1)dt)/((a^2 t^2 +b^2 )^2 ))    = (π/a^2 )∫_0 ^( ∞)  ((a^2 t^2 +a^2 )/((a^2 t^2 +b^2 )^2 ))dt    =  (π/a^2 ) ∫_(0 ) ^( ∞)    (( (a^2 −b^2 )+(a^2 t^2 +b^2 ))/((a^2 t^2 +b^2 )^2 ))dt   = (π/a^2 )∫_0 ^( ∞)  (((a^2 −b^2 ))/((a^2 t^2 +b^2 )^2 ))dt+ (π/a^2 )∫_0 ^∞  (1/(a^2 t^2 +b^2 ))dt   = ((π(a^2 −b^2 ))/a^2 )∫_0 ^( ∞)  (1/((a^2 t^2 +b^2 )^2 ))dt+(π/(a^2 .a^2 ))∫_0 ^∞ (1/((t)^2 +((b/a))^2 ))dt    here, I= I_1 +I_2     I_1 =  ((π(a^2 −b^2 ))/a^2 )∫_0 ^( ∞)  (1/((a^2 t^2 +b^2 )^2 ))dt       Put, at= b tanθ               adt= b sec^2 θdθ         dt= (b/a) sec^2 θdθ    If  t=∞ ⇒ θ= (π/2)    If t=0 ⇒ θ=0    ((π(a^2 −b^2 ))/a^2 )(b/a)∫_0 ^( (π/2))  ((sec^2 θdθ)/((b^2 tan^2 θ+b^2 )^2 ))    = ((π(a^2 −b^2 )b)/a^3 )∫_0 ^(π/2)  ((sec^2 θdθ)/(b^4 .sec^4 θ))   = ((π(a^2 −b^2 ))/(a^3 b^3 ))∫_0 ^(π/2) cos^2 θdθ     cos2θ= 1−2cos^2 θ⇒ cos^2 θ =((1−cos2θ)/2)      [ (1/2)∫(1−cos2θ)dθ]_0 ^(π/2)     = [ (1/2)(θ − ((sin2θ)/2))]_0 ^(π/2) ⇒ [(1/2)((π/2)−((sin2×(π/2))/2))−0]    = (π/4)   I_1 =  ((π(a^2 −b^2 ))/(a^3 b^3 ))×(π/4)   Again  for     I_2 = (π/a^4 )[ (1/(b/a)) tan^(−1)  (t/(b/a))]_0 ^∞      = (π/a^4 )[ (a/b)tan^(−1) ((at)/b)]_0 ^∞      = (π/a^4 )[ (a/b)×(π/2)−0]    = (π^2 /(2a^3 b))   Now complete solution    I= I_1 +I_2     =  ((π^2 (a^2 −b^2 ))/(4a^3 b^3 ))+ (π^2 /(2a^3 b))   = (π^2 /(2a^3 b))(((a^2 −b^2 )/(2b^2 ))+1)   = (π^2 /(2a^3 b))(((a^2 −b^2 +2b^2 )/(2b^2 )))   = ((π^2 (a^2 +b^2 ))/(4a^3 b^3 ))  //.    Which is required to proof.

1.I=0πxdx(a2sin2x+b2cos2x)2.....(i)=0π(πx)dx{a2sin2(πa)+b2cos2(πa)}2I=0π(πx)dx(a2sin2x+b2cos2x)2....(ii)Added(i)+(ii)2I=0ππ(a2sin2x+b2cos2x)2dxI=π2×20π21(a2sin2x+b2cos2x)2dx=π0π21(a2sin2x+b2cos2x)2dxDividenumerator&denominatorby1cos4x=π0π2sec4xdx{1cos2x(a2sin2x+b2cos2x)}2=π0π2(tan2x+1)sec2xdx(a2tan2x+b2)2Puttanx=tsec2xdx=dtIfx=π2t=Ifx=0t=0Now,π0(t2+1)dt(a2t2+b2)2=πa20a2t2+a2(a2t2+b2)2dt=πa20(a2b2)+(a2t2+b2)(a2t2+b2)2dt=πa20(a2b2)(a2t2+b2)2dt+πa201a2t2+b2dt=π(a2b2)a201(a2t2+b2)2dt+πa2.a201(t)2+(ba)2dthere,I=I1+I2I1=π(a2b2)a201(a2t2+b2)2dtPut,at=btanθadt=bsec2θdθdt=basec2θdθIft=θ=π2Ift=0θ=0π(a2b2)a2ba0π2sec2θdθ(b2tan2θ+b2)2=π(a2b2)ba30π2sec2θdθb4.sec4θ=π(a2b2)a3b30π2cos2θdθcos2θ=12cos2θcos2θ=1cos2θ2[12(1cos2θ)dθ]0π2=[12(θsin2θ2)]0π2[12(π2sin2×π22)0]=π4I1=π(a2b2)a3b3×π4AgainforI2=πa4[1batan1tba]0=πa4[abtan1atb]0=πa4[ab×π20]=π22a3bNowcompletesolutionI=I1+I2=π2(a2b2)4a3b3+π22a3b=π22a3b(a2b22b2+1)=π22a3b(a2b2+2b22b2)=π2(a2+b2)4a3b3//.Whichisrequiredtoproof.

Commented by Mr.D.N. last updated on 22/May/20

Thanks for all of you to presents your ability it's really I appreciate.. ������

Answered by niroj last updated on 22/May/20

 2. Sol^n :     If m be a mass of the lamania,then   m= ∫_(R ) ∫ϱ dx dy = ∫_R ∫ϱx dx dy    where R being the circle      x^2 +y^2 −2ax =0  We know polar transformation    x= r cosθ ,  y= r sinθ    then equation of the circle be comes   x^2 +y^2 = 2ax  i.e. r^2 =2ar cosθ i.e.  r=2a cosθ    Also, dxdy = r dr dθ     The bounded of R of the circle can be described by−(π/2)≤ θ≤(π/2) and     0≤r ≤2a cosθ    ∴  m = ∫_(θ=((−π)/2)) ^(π/2)  ∫_0 ^( 2a cosθ) r cosθ r dr dθ      = ∫_(−π/2) ^(π/2)  cosθ ((r^2 /3))_0 ^(2a cosθ) dθ     = ∫_(−π/2) ^( π/2)  ((8a^3 )/3).cos^4 θ dθ    = ((16 a^3 )/3) ∫_0 ^( π/2)  cos^4 θdθ     by using gamm function    = ((16a^3 )/3)((  ⌈4 +(1/2).⌈(1/2))/(2⌈3))   = ((16a^3 )/3). (((3/2).(1/2)(√π) . (√π))/(2.2.1))   = ((16 a^3 )/3).(3/4).(1/4).π   = π a^3    ∴ It′s mass = π a^3  //.

2.Soln:Ifmbeamassofthelamania,thenm=Rϱdxdy=RϱxdxdywhereRbeingthecirclex2+y22ax=0Weknowpolartransformationx=rcosθ,y=rsinθthenequationofthecirclebecomesx2+y2=2axi.e.r2=2arcosθi.e.r=2acosθAlso,dxdy=rdrdθTheboundedofRofthecirclecanbedescribedbyπ2θπ2and0r2acosθm=θ=π2π202acosθrcosθrdrdθ=π/2π/2cosθ(r23)02acosθdθ=π/2π/28a33.cos4θdθ=16a330π/2cos4θdθbyusinggammfunction=16a334+12.1223=16a33.32.12π.π2.2.1=16a33.34.14.π=πa3Itsmass=πa3//.

Commented by Mr.D.N. last updated on 22/May/20

what nice outstanding��������

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