二重積分の問題です

質問

問題を写真の通りです、詳しい解答方法をお願いします
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条件を補足します
y=f(x) (0<x<2π)

回答No.2

jcpmutura · ベストアンサー

f"(x)はf'(x)のxによる微分であってtによる微分ではありません
f"(x)=(d/dx)f'(x)≠(d/dt)f'(x)=-1/(1-cos(t))

0≦t≦2π
x(t)=t-sint
y(t)=1-cost
y(t)=f(x(t))
1.
x_0=x(π/3)
a_0=f(x_0)=f(x(π/3))=y(π/3)=1-cos(π/3)=1-1/2=1/2
x'(t)=1-cost
y'(t)=sint
f'(x)=y'(t)/x'(t)=sint/(1-cost)
x'(π/3)=1-cos(π/3)=1-1/2=1/2
y'(π/3)=sin(π/3)=√3/2
a_1=f'(x_0)=y'(π/3)/x'(π/3)=√3
(d/dt){f'(x)}
=cost/(1-cost)-(sint)^2/(1-cost)^2
=1/(cost-1)
f"(x)=(d/dt){f'(x)}/x'(t)=1/(cost-1)/(1-cost)=-1/(1-cost)^2
f"(x_0)=f"(x(π/3))=-1/{1-cos(π/3)}^2=-4
a_2=f"(x(π/3))/2=-4/2=-2

∴
a_0=1/2
a_1=√3
a_2=-1/2

2.
D={(x,y)|x=t-sint,0≦y≦1-cost,0≦t≦2π}
∬Ddxdy
=∫_{0～2π}f(x)dx
=∫_{0～2π}y(t)x'(t)dt
=∫_{0～2π}{(1-cost)^2}dt
=∫_{0～2π}{1-2cost+(cost)^2}dt
=∫_{0～2π}[1+{1+cos(2t)}/2]dt
=∫_{0～2π}[3/2+{cos(2t)}/2]dt
=(3/2)∫_{0～2π}dt
=2π*3/2
=3π

∬Dxdxdy
=∫_{0～2π}xf(x)dx
=∫_{0～2π}x(t)y(t)x'(t)dt
=∫_{0～2π}{(t-sint)(1-cost)^2}dt
=∫_{0～2π}[(t-sint){1-2cost+(cost)^2}]dt
=∫_{0～2π}[t-2tcost+t{1+cos(2t)}/2-sint+sin(2t)-sint{1+cos(2t)}/2]dt
=∫_{0～2π}[t-2tcost+t{1+cos(2t)}/2-sintcos(2t)}/2]dt
=∫_{0～2π}[3t/2-2tcost+tcos(2t)/2+{sin(t)-sin(3t)}/4]dt
=∫_{0～2π}[3t/2-2tcost+tcos(2t)/2]dt
=[3t^2/4-2tsint+tsin(2t)/4]_{0～2π}+∫_{0～2π}(2sint-sin(2t)/4)dt
=3π^2

∬Dydxdy
=∫_{0～2π}{f(x)}^2dx
=∫_{0～2π}[{y(t)}^2]x'(t)dt
=∫_{0～2π}{(1-cost)^3}dt
=∫_{0～2π}(1-cost){1-2cost+(cost)^2}dt
=∫_{0～2π}(1-cost){(3/2)-2cost+{cos(2t)}/2}dt
=∫_{0～2π}[(3/2)-(7/2)cost+2(cost)^2+{cos(2t)}/2-{costcos(2t)}/2]dt
=∫_{0～2π}[(5/2)+{3cos(2t)}/2-{costcos(2t)}/2]dt
=∫_{0～2π}[(5/2)-(15/2)cost-{cos(3t)}/4]dt
=(5/2)∫_{0～2π}dt
=5*2π/2
=5π

回答No.3

jcpmutura

#2です
a_2=-2
に訂正します

0≦t≦2π
x(t)=t-sint
y(t)=1-cost
y(t)=f(x(t))
1.
x_0=x(π/3)
a_0=f(x_0)=f(x(π/3))=y(π/3)=1-cos(π/3)=1-1/2=1/2
x'(t)=1-cost
y'(t)=sint
f'(x)=y'(t)/x'(t)=sint/(1-cost)
x'(π/3)=1-cos(π/3)=1-1/2=1/2
y'(π/3)=sin(π/3)=√3/2
a_1=f'(x_0)=y'(π/3)/x'(π/3)=√3
(d/dt){f'(x)}
=cost/(1-cost)-(sint)^2/(1-cost)^2
=1/(cost-1)
f"(x)=(d/dt){f'(x)}/x'(t)=1/(cost-1)/(1-cost)=-1/(1-cost)^2
f"(x_0)=f"(x(π/3))=-1/{1-cos(π/3)}^2=-4
a_2=f"(x(π/3))/2=-4/2=-2

∴
a_0=1/2
a_1=√3
a_2=-2

2.
D={(x,y)|x=t-sint,0≦y≦1-cost,0≦t≦2π}
∬Ddxdy
=∫_{0～2π}f(x)dx
=∫_{0～2π}y(t)x'(t)dt
=∫_{0～2π}{(1-cost)^2}dt
=∫_{0～2π}{1-2cost+(cost)^2}dt
=∫_{0～2π}[1+{1+cos(2t)}/2]dt
=∫_{0～2π}[3/2+{cos(2t)}/2]dt
=(3/2)∫_{0～2π}dt
=2π*3/2
=3π

∬Dxdxdy
=∫_{0～2π}xf(x)dx
=∫_{0～2π}x(t)y(t)x'(t)dt
=∫_{0～2π}{(t-sint)(1-cost)^2}dt
=∫_{0～2π}[(t-sint){1-2cost+(cost)^2}]dt
=∫_{0～2π}[t-2tcost+t{1+cos(2t)}/2-sint+sin(2t)-sint{1+cos(2t)}/2]dt
=∫_{0～2π}[t-2tcost+t{1+cos(2t)}/2-sintcos(2t)}/2]dt
=∫_{0～2π}[3t/2-2tcost+tcos(2t)/2+{sin(t)-sin(3t)}/4]dt
=∫_{0～2π}[3t/2-2tcost+tcos(2t)/2]dt
=[3t^2/4-2tsint+tsin(2t)/4]_{0～2π}+∫_{0～2π}(2sint-sin(2t)/4)dt
=3π^2

∬Dydxdy
=∫_{0～2π}{f(x)}^2dx
=∫_{0～2π}[{y(t)}^2]x'(t)dt
=∫_{0～2π}{(1-cost)^3}dt
=∫_{0～2π}(1-cost){1-2cost+(cost)^2}dt
=∫_{0～2π}(1-cost){(3/2)-2cost+{cos(2t)}/2}dt
=∫_{0～2π}[(3/2)-(7/2)cost+2(cost)^2+{cos(2t)}/2-{costcos(2t)}/2]dt
=∫_{0～2π}[(5/2)+{3cos(2t)}/2-{costcos(2t)}/2]dt
=∫_{0～2π}[(5/2)-(15/2)cost-{cos(3t)}/4]dt
=(5/2)∫_{0～2π}dt
=5*2π/2
=5π

回答No.1

info222_

1
t=π/3のときx0=π/3-sin(π/3)=π/3-√3/2
y(t)=f(x(t))
f(x0)=y(π/3)=1-cos(π/3)=1/2
y'=sin(x)=f '(x) x'=f '(x) y
f '(x)=y'/y=sin(t)/(1-cos(t))
f '(x0)=sin(π/3)/(1-cos(π/3))=√3
f ''(x)=(cos(t)(1-cos(t))-sin(t)sin(t))/(1-cos(t))^2
=(cos(t)-1)/(1-cos(t))^2=-1/(1-cos(t))
f ''(x0)=-1/(1-cos(π/3))=-2
∴a0=f(x0)= 1/2, a1=f '(x0)=√3, a2=f ''(x0)/2!= -2/2= -1

2
I0=∫∫[D] dxdy=∫[0,2π] f(x) dx
x:0→2πのときt=0→2πなので
=∫[0,2π] y(t) (dx/dt)dt
=∫[0,2π] (1-cos(t))^2 dt
=∫[0,2π] (1-2cos(t)+(1/2)(1+cos(2t))) dt
=∫[0,2π] (3/2) dt
=3π

Ix=∫∫[D] xdxdy= ∫[0,2π] xydx
=∫[0,2π] xy (dx/dt) dt
x:0→2πのときt=0→2πなので
=∫[0,2π] (t-sin(t)) (1-cos(t))^2 dt
=∫[0,2π] (t-sin(t)) (1-2cos(t)+(1/2)(1+cos(2t))) dt
三角関数の１周期の整数倍の区間の積分は0になるから
=∫[0,2π] t(1-2cos(t)+(1/2)(1+cos(2t))) dt
部分積分する。三角関数の１周期の整数倍の区間の積分は0になるから
=[t((3/2)t-2sin(t)+(1/4)sin(2t))][0,2π]-∫[0,2π] (3/2)t dt
=6π^2-[(3/4)t^2][0,2π]
=3π^2

Iy=∫∫[D] ydxdy= ∫[0,2π] y^2 dx
=∫[0,2π] y^2 (dx/dt) dt
x:0→2πのときt=0→2πなので
=∫[0,2π] (1-cos(t))^3 dt
=∫[0,2π] (1-3cos(t)+3(cos(t))^2-(cos(t))^3) dt
=∫[0,2π] (1-3cos(t)+(3/2)(1+cos(2t))-(1/4)(3cos(t)+cos(3t))) dt
三角関数の１周期の整数倍の区間の積分は0になるから
=∫[0,2π] (5/2) dt
=5π