重積分・2変数積分

著者:Mathcot.H.I.
初版: 2007.05.03
Update:2014.01.19

直交座標系
(x,y,z), z=f(x,y)
定積分:∬D f(x,y) dxdy

不定積分:∬ f(x,y) dxdy

極座標系:円柱座標
(r,θ,z), r=√(x2+y2), θ=θ, z=z
0≦θ<2π
x=r cosθ, y=r sinθ, z=z

dxdy=rdrdθ
dxdydz=rdrdθdz


極座標系:球座標系
(r,θ,φ), r=√(x2+y2+z2), θ=θ,φ=φ
φ=tan-1 (z/√(x2+y2)=tan-1 (z/√(r2-z2)
x=r sinθcosφ, y=r sinθsinφ, z=r cosθ

dxdy=r drdθ
dxdydz=|J|drdθdφ=r2sinθdrdθdφ





定積分
 
f(x,y)
F(x,y)=∫∫D f(x,y)dxdy (積分定数は省略)
x/(x+y)2 






参考URL
[1]基 本的な積分公式
[2]公 式(積分)(東海大学)
[3] 不定積分サイト:integrals.wolfram.com/index.jsp
[4] Common Integrals
[5] COMMON SUBSTITUTIONS
[6] Common Integrals INTEGRALS CONTAINING ax+b
[7] INTEGRALS CONTAINING THE SQUARE ROOT OF ax+b
[8] INTEGRALS CONTAINING ax+b AND px+q
[9] Tables
[10]
[11]

定積分
                    
f(x,y),[積分範囲下限,上限
積分値                
[演習1] f(x,y)=x,
  D={(x,y)|x2+y2≦2y,y≦x}の時
D f(x,y)dxdyを求めよ。
TS-01.PNG
[解答] ∬D xdxdy=∫[0,1]dy∫[y,√{1-(y-1)2}] xdx
=∫[0,1] (1/2)[{1-(y-1)2}-y2]dy
=∫[0,1] (y-y2)dy=[(1/2)y2-(1/3)y3]|(0,1)
=(1/2)-(1/3)=1/6
[別解] x=r cos(t), y=r sin(t)とおけば、領域Dは
r^2≦r sin(t)
0≦t≦π/4,r≧0
D→D'={(r,t)|r≦2sin(t),0≦t≦π/4}
∫[t:0,π/4]∫[r:0,2sin(t)] (r^2)cos(t)drdt
=∫[t:0,π/4] cos(t)∫[r:0,2sin(t)] (r^2)drdt
=∫[t:0,π/4] cos(t)[(r^3)/3]|(0,2sin(t))dt
=(1/3)∫[t:0,π/4] cos(t)*8(sin(t))^3 dt
=(8/3)[{(sin(t))^4}/4]|(0,π/4)
=(2/3)(1/√2)^4=1/6
[演習2] f(x,y)=(x+y)4,
  D={(x,y)||x|+|y|≦1}の時
D f(x,y)dxdyを求めよ。
TS-02.PNG
[解答] ∫D (x+y)4dxdy
u=x+y,v=y-xと変換
|J|=1/2 , dxdy=(1/2)dudv
(x+y)4=u4
領域D → D'={(u,v)|-1≦u≦1,-1≦v≦1}
積分は
∫∫D' (u4)*(1/2)du}dv
=∫[-1,1]{∫[-1,1] (u4)*(1/2)du}dv
=2{∫[-1,1] (u4)*(1/2)du}=[u4/5]|(-1,1)
=2/5
[演習3] f(x,y)=ln(x)/y2,
 D={(x,y)|1≦y≦x≦2}の時
D f(x,y)dxdyを求めよ。
TS-03.PNG
[解答] ∬D ln(x)/y2 dxdy
=∫12ln(x)∫1x y-2dy dx
=∫12ln(x){1-(1/x)} dx
=2ln(2)-1-(1/2){ln(2)}2
[演習4] f(x,y)=sin(2x+y),
 D={(x,y)|0≦x≦π/2,x≦y≦2x}の時
D f(x,y)dxdyを求めよ。
TS-04.PNG
[解答] ∬D sin(2x+y)dxdy(D={(x,y)|0≦x≦π/2,x≦y≦2x})
=∫[0→π/2]{∫[x→2x] sin(2x+y)dy}dx
=∫[0→π/2] [-cos(2x+y)]|(x→2x)dx
=∫[0→π/2](-cos4x+cos3x)dx
=[-(1/4)sin4x+(1/3)sin3x]|(0→π/2)
=-1/3
[演習5] f(x,y)=|sin(2x+y)|,
 D={(x,y)|0≦x≦π/2,x≦y≦2x}の時
D f(x,y)dxdyを求めよ。
TS-05.PNG
[解答] ∬D |sin(2x+y)|dxdy(D={(x,y)|0≦x≦π/2,x≦y≦2x})
=∫[0→π/4]{∫[x→2x] sin(2x+y)dy}dx+∫[π/4→π/3]{∫[x→π-2x] sin(2x+y)dy}dx-∫[π/4→π/3]{∫[π-2x→2x] sin(2x+y)dy}dx-∫[π/3→π/2]{∫[x→2x] sin(2x+y)dy}dx
=(√2)/6+{(π/12)-(√2)/6}+{(π/12)-(√3)/8}+{(√3)/8+(1/3)}
=(π/6)+(1/3)

[別解] u=2x-y,v=2x+yとおくと∂(x,y)/∂(u,v)=1/4
dxdy=(1/4)dudv
 D1={(u,v)|0≦u+v≦2π,0≦v≦π,(u+v)≦2(v-u)≦2(u+v)}
  ={(u,v)|0≦3u≦v≦π}
 D2={(u,v)|0≦u+v≦2π,π≦v≦3π/2,(u+v)≦2(v-u)≦2(u+v)}
   ={(u,v)|π≦v,0≦3u≦v≦3π/2}
 D3={(u,v)|0≦u+v≦2π,3π/2≦v≦2π,(u+v)≦2(v-u)≦2(u+v)}
   ={(u,v)|3π/2≦v≦2π,0≦u≦2π-v}
D |sin(2x+y)|dxdy
=∬D1sin(v)(1/4)dudv-∬D2sin(v)(1/4)dudv-∬D32sin(v)(1/4)dudv
=(1/4)∫[0→π]{sin(v)∫[0→v/3] du}dv
 -(1/4)∫[π→3π/2]{sin(v)∫[0→v/3] du}dv
 -(1/4)∫[3π/2→2π]{sin(v)∫[0→2π-v] du}dv
=(π/12)+{(π+1)/12}+(1/4)=(π/6)+(1/3)

[演習6] I=∬D x2dxdy,
 D={(x,y)|(x2/a2)+(y2/b2)≦1,a>0,b>0} を求めよ。
[解答] x=arcos(t),y=brsin(t)とおく と,∂(x,y)/∂(r,t)=acos(t)*brcos(t)+arsin(t)*bsin(t)=abr
dxdy=abrdrdt,E={(r,t)|0≦r≦1,-π≦t≦π}
I=∬E a3br3cos2(t)drdt
=(a3b/2)∫[t:-π,π] (1+cos(2t))dt∫[r:0,1]r3dr
=(a3b)∫[t:0,π] (1+cos(2t))dt[r4/4]|(0,1)
=(a3b/4)[(t+sin(2t)/2)]|(0,π)
=a3bπ/4
[演習7] I=∬D (x+y)4dxdy,
 D={(x,y)|(x+y)2+y2≦1}を求めよ。
[解答] u=x+y,v=yとおくと∂(x,y)/∂(u,v)=1
dxdy=dudv
 D→E={(u,v)|0≦u2+v2≦1}
I=∬E u4dudv
=2∫[v:0,1] dv ∫[u:-√(1-v2),√(1-v2)] u4du
=(4/5)∫[v:0,1] [u5]|[u:0,√(1-v2)] dv
=(4/5)∫[v:0,1] (1-v2)5/2dv
=(4/5)(1/48)[v(1-v2)1/2(8v4-26v2+33)+15sin-1v]|[v:0,1]
=(4/5)(5/32)π=π/8
[別解] u=x+y,v=yとおくと∂(x,y)/∂(u,v)=1
dxdy=dudv
 D→E={(u,v)|0≦u2+v2≦1}
I=∬E u4dudv
u=r cosθ,v=r sinθと置くとdudv=rdrdθ
 E→F={(r,θ)|0≦r≦1,-π≦θ≦π}
I=∬F r5(cosθ)4drdθ
=2∫[θ:0,π] (cosθ)4dθ∫[r:0,1] r5dr
=(2/6)[r6]|[r:0,1] ∫[θ:0,π] (cosθ)4
=(1/3)∫[θ:0,π] (cosθ)4
=(1/24)[cosθsinθ(2cos2θ+3)+3θ]|[θ:0,π]
=(1/24)3π=π/8

[演習8] ∬D exp(2x2+2xy+y2)dxdy,
  D={(x,y}|全実数範囲}を求めよ。
[解答]
∫[x:-∞,∞]∫[y:-∞,∞] c e^-(2x^2+2xy+y^2)dxdy=1
I=∫[x:-∞,∞]∫[y:-∞,∞] e^-(2x^2+2xy+y^2)dxdy=1/c
=∫[y:-∞,∞]e^(-y^2)dy∫[x:-∞,∞] e^-(2x^2+2xy)dx
=∫[-∞,∞]e^(-y^2)dy*(√(π/2))*exp((y^2)/2)
=(√(π/2))*∫[-∞,∞] exp(-(y^2)/2)dy
=(√(π/2))*√(2π)=π

[演習9] I=∬D (x2+y2)/(x+y)^2 dxdy,
  D={(x,y)|1≦x+y≦2,x≧0,y≧0}を求めよ。
[解答]
x=(u+v)/2, y=(v-u)/2とおくと
|J|=|∂(x,y)/∂(u,v)|
=|(∂x/∂u)(∂y/∂v)-(∂x/∂v)(∂y/∂u)|
=|(1/2)(1/2)-(1/2)(-1/2)|=1/2
dxdy=|J|dudv=(1/2)dudv
(x2+y2)/(x+y)2=(1/4)((u+v)2+(v-u)2)/v2=(1/2)(u2+v2)/v2
D'={(u,v)|1≦v≦2,u+v≧0,v-u≧0}
  ={(u,v)|1≦v≦2,v≧-u,v≧u}
D"={(u,v)|1≦v≦2,v≧u≧0}
I=∬D' (1/4)(u2+v2)/v2 dudv
 =(1/2)∬D" (u2+v2)/v2 dudv
 =(1/2){∫[0,2]du∫[u,2] (u^2/v^2)+1 dv
  -∫[0,1]du∫[u,1](u^2/v^2)+1dv }
 =(1/2){∫[0,2] [-(u^2/v)+v] [v=u,2] du
  -∫[0,1] [-(u^2/v)+v] [v=u,1] du }
 =(1/2){∫[0,2] [-(u^2/2)+2] du
  -∫[0,1] [-(u^2)+1] du }
 =(1/2){[-(u^3/6)+2u] [u=0,2]
  -[-(u^3/3)+u][u=0,1]}
 =(1/2){-(4/3)+4-(-(1/3)+1)}
 =1
[演習10] 2本の半径a(>0) の直交円柱の共通部分の体積Vを求めよ。

 V=∫∫∫[D] dxdydz . D:{(x,y,z)|x2+z2≦a2,y2+z2≦a2}
2entyu-kantu-tai.PNG
[解答]
V =16∬[D'] z dxdydz , D':{(x,y,z)|0≦y≦x≦a,x2+z2≦a2,0≦z}
 =16∬[E] √(a2-x2) dxdy , E:{(x,y)|0≦y≦x≦a}
 =16[0,a] dx [0,x) √(a2-x2) dy
 =16[0,a] x √(a2-x2) dx
 =16 [-(1/3)(a2-x2)3/2] [0,a]
 = (16/3) a3
 
演習11] f(x,y)=(x2+y^2) のとき

 V=∫∫D f(x,y) dxdy , D={(x,y)| 0≦x≦√2, 0≦y≦√2,} を求めよ。
[解答]
積分領域Dと被積分関数のx,yの対称性から
積分領域D={(x,y)|0≦x≦2、0≦y≦2}
を2分割して
積分領域E1={(x,y)|0≦y≦x≦2}
とE2={(x,y)|0≦x≦y≦2}
とすれば
I=∫∫(x^2+y^2) dxdy
 =∫∫E1 (x^2+y^2) dxdy+∫∫E2 (x^2+y^2)dxdy
 =2∫∫E1 √(x^2+y^2) dxdy
ここで極座標変換して
  x=rcos(t),y=rsin(t)
  E1⇒F:{(r,t)|.0≦r≦√2/cos(t), 0≦t<π/4}
  √(x^2+y^2)dxdy=r^2 drdt
なので
I=2∫∫[F] r^2 drdt
=2∫[0,π/4] dt∫[0,√2/cos(t)] r^2 dr
=2∫[0,π/4] {[(1/3)r^3][0,√2/cos(t)]}dt
=(2/3)∫[0,π/4]{√2/cos(t)}^3 dt
=((4√2)/3)∫[0,π/4] cos(t)/(1-(sin(t))^2)^2 dt
sin(t)=u(0≦u<1/√2)とおいて置換積分
  cos(t)dt=du, cos(t)/(1-(sin(t))^2)^2 dt=1/(1-u^2)^2 du
なので
I=((4√2)/3)∫[0,1/√2] 1/(u^2-1)^2 du

部分分数分解して
=((√2)/3)∫[0,1/√2] {1/(u+1) +1/(1-u) +1/(1-u)^2 +1/(u+1)^2} du
=((√2)/3)[ln((u+1)/(1-u)))-1/(u-1)-1/(u+1)][0,1/√2]
=((√2)/3){ln((1+√2)/(√2-1)) -√2/((1/2)-1)}
=((√2)/3){2ln(1+√2) +2√2}
=((2√2)/3)ln(1+√2) +(4/3)
[演習12]  f(x,y)=xy2/√(a2-x2) (a>0) のとき

 V=∫∫[D] f(x,y)dxdy , D:{(x,y,z)|x2+y2≦a2, 0≦x} を求めよ。



[解答] 
∬ f(x,y) dxdy
=∫[-a→a] y2 dy∫[0→√(a2-y2)] x/√(a2-x2)dx
=∫[-a→a] y2 dy[-√(a2-x2)][0→√(a2-y2)]
√y2=|y|なので
=∫[-a→a] y2 (a-|y|) dy
 |y|の絶対値をはずすためyの積分範囲を分けると
=∫[-a→0] (ay2+y3)dy+∫[0→a] (ay2-y3)dy
=[ay3/3+y4/4][-a→0] + [ay3/3-y4/4][0→a]
=a4/3 -a4/4 +a4/3 -a4/4
=a4 /6
[別解]
∬ f(x,y) dxdy
逐次積分に直して
=∫[0→a] x/√(a2-x2) dx ∫[-√(a2-x2)→√(a2-x2)] y2 dy
=2∫[0→a] x/√(a2-x2) dx ∫[0→√(a2-x2)] y2 dy
=2∫[0→a] x/√(a2-x2) dx [ y3/3][0→√(a2-x2)]
=(2/3) ∫[0→a] x(a2-x2) dx
=(2/3) [ a2x2/2 -x4/4 ][0→a]
=(2/3) (a4/2 -a4/4)
=a4 /6
[演習13]  f(x,y)=1/x2 のとき

 V=∫∫[D] f(x,y) dxdy , D:{(x,y) |1/x≦y≦x, 1≦x≦2} を求めよ。

JSB1226a.PNG

JSB1226b.PNG

[解答] 図1参照
∫∫[D] 1/x^2dxdy、D={(x,y)|1/x≦y≦x,1≦x≦2}
逐次積分に直すと
=∫[1→2] (1/x2)dx∫[1/x→x] dy
y(>0)で積分すると
=∫[1→2] (1/x2)(x-1/x) dx
=∫[1→2] (1/x)-(1/x3) dx
x(>0)で積分する。ln()を自然対数として
=[ln(x)+(1/(2x2))][1→2]
=log(2)-3/8

















[別解] 図2参照
∫∫[D] 1/x^2dxdy、D={(x,y)|1/x≦y≦x,1≦x≦2}
逐次積分に直すと
=∫[1/2→1]dy ∫[1/y→2] (1/x2)dx+∫[1→2]dy ∫[y→2] (1/x2)dx
xで積分すると
=∫[1/2→1]dy [-1/x][1/y→2]+∫[1→2]  dy [-1/x][y→2]
=∫[1/2→1] (y-1/2) dy+∫[1→2]  (1/y-1/2) dy
y(>0)で積分すると
=[y2/2-y/2][1/2→1]+[ln(y)-y/2][1→2]
=1/2-1/8+1/4-1/2+ln(2) +1/2-1
=ln(2)-3/8
[演習14]  f(x,y)=xy2 のとき

 V=∫∫[D] f(x,y) dxdy , D:{(x,y) | x2+y2≦a2,x≧0} (a>0) を求めよ。


[解答]
∫∫[D] xy2dxdy、D={(x,y)|x2+y2≦a2,x≧0} (a>0)
逐次積分に直すと
=2∫[0→a] xdx∫[0→√(a2-x2)] y2 dy
yで積分すると
=2∫[0→a] xdx [(1/3)y3][0→√(a2-x2)]
=(2/3)∫[0→a] x(a2-x2)3/2 dx
合成関数の積分公式を適用して
=(2/3)[-(1/5)(a2-x2)5/2][0→a]
=(2/15)a5 
[演習15] 

 V=∫∫[D]  dxdydz , D:{(x,y,z) | x2+y2≦4, 0≦z≦2-y}  を求めよ。

E14ENTYU.PNG
[解答]
V=∫∫∫{0≦z≦2-y,x2+y2≦4} dxdydz
 =2∫∫{x2+y2≦4} (2-y)dxdy
 =2∫[-2→2]dy∫[0→√(4-y2)] (2-y)dx
または
 =2∫[0→2] dx∫[-√(4-x2)→√(4-x2)] (2-y)dy
[解1]
V=2∫[-2→2] dy∫[0→√(4-y2)] (2-y)dx
 =2∫[-2→2]  (2-y)√(4-y2) dy
 =8∫[0→2] √(4-y2) dy
 =8∫[0→2] (4-y2)1/2 dy
 =8{[y(4-y2)1/2][0→2]+∫[0→2] y2(4-y2)-1/2 dy}
 =8∫[0→2] y2(4-y2)-1/2 dy
 =8∫[0→2] (4-(4-y2))(4-y2)-1/2 dy
 =32∫[0→2] (4-y2)-1/2 dy -8∫[0→2] (4-y2)1/2 dy
 =16∫[0→2] (4-y2)-1/2 dy
 =16[sin-1(y/2)][0→2]
 =16π/2
 =8π

[解2]
V=2∫[0→2] dx∫[-√(4-x2)→√(4-x2)] (2-y)dy
 =4∫[0→2] dx∫[0→√(4-x2)] 2dy
 =4∫[0→2] 2(4-x2)1/2 dx
 =8∫[0→2] (4-x2)1/2 dx
 =8∫[0→2] (4-x2)1/2 dx
 =8{[y(4-x2)1/2][0→2]+∫[0→2] x2(4-x2)-1/2 dx}
 =8∫[0→2] x2(4-x2)-1/2 dx
 =8∫[0→2] (4-(4-x2))(4-x2)-1/2 dx
 =32∫[0→2] (4-x2)-1/2 dx -8∫[0→2] (4-x2)1/2 dx
 =16∫[0→2] (4-x2)-1/2 dx
 =16[sin-1(x/2)][0→2]
 =16π/2
 =8π
[演習16]
V=∫∫D x/(x+y)2 dxdy, D={(x,y)|1≦x≦2,0≦y≦1} を求めよ。

[解答]
V=∫[1,2] xdx ∫[0,1] (y+x)^(-2)dy
=∫[1,2] xdx [-(y+x)^(-1)][y:0,1]
=∫[1,2] x [x^(-1)-(1+x)^(-1)] dx
=∫[1,2] [1 -x/(1+x)] dx
=∫[1,2] 1/(1+x) dx
=[ln(x+1)][1,2]
=ln(3)-log(2)
=ln(3/2)
[演習17]
V=∫∫D (|x|+|y|) dxdy, D={(x,y)| |x|+|y|≦1} を求めよ。
[解答]
被積分関数(|x|+|y|)および積分領域がx軸およびy軸に対称なので
積分領域を
 D→ E:{(x,y)| x+y≦1,x≧1,y≧1}
に変換すれば、積分領域Eでは絶対値を外すことができ、
積分領域Dにおける積分は、積分領域Eにおける積分の4倍になるから
V=4∫E (x+y)dxdy, E:{(x,y)| x+y≦1,x≧1,y≧1}
=4∫[0→1]dx∫[0→1-x](x+y)dy
=4∫[0→1]dx [xy+y2/2][0→1-x]
=4∫[0→1] [x(1-x)+(1-x)2/2] dx
=4[x2/2 -x3/3 -(1-x)3/6][0→1]
=4(1/2 -1/3 +1/6)
=4/3
[演習18]2平面 z=0, z=2-y と円柱面 x2+y2=4  で囲まれる部分の体積Vを求めよ。
 V=∫∫∫D dxdydz, D={(x,y,z)| 0≦z≦2-y, x2+y2≦4}
[解答]
累次(逐次)積分で表すと
 V=∫∫∫{0≦z≦2-y,x2+y2≦4} dxdydz
 =2∫∫{x2+y2≦4} (2-y)dxdy
 =2∫[-2→2]dy ∫[0→√(4-y2)] (2-y) dx
 =2∫[-2→2] (2-y)√(4-y2) dy
 =8∫[0→2] √(4-y2) dy
y=2sin(t)とおいて置換積分
 =8∫[0→π/2] 2√(1-sin2(t)) 2cos(t)dt
 =32 ∫[0→π/2] cos2(t) dt
 =16 ∫[0→π/2] 1+cos(2t) dt
 =16(π/2)=8π

[別解]
 V=2∫[0→2] dx ∫[-√(4-x2)→√(4-x2)] (2-y)dy
 =2∫[0→2] dx 2∫[0→√(4-x2)] 2dy
 =8∫[0→2] √(4-x2) dx
 x=2sin(t)とおいて置換積分
 x:0→2のとき t:0→π/2
 dx=2cos(t)dt
 V=8∫[0→π/2] 4cos2(t) dt
 =16∫[0→π/2] 1+cos(2t) dt
 =16(π/2)=8π
参考URL
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改訂履歴
Update: 2007.6.20
Update:2008.01.27
Update:2008.11:21
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Update:2012.12:26
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