急いでます。.重積分の問題です

質問

急いでます。.重積分の問題です

(1) ∫∫√（xy-x^2）dxdy {(x,y)|0<x<y<2x<2}
(2)　曲面bz= ｘ^2＋ｙ^2(b>0)と円柱面ｘ^2＋ｙ^2=ax(a>0)と平面：ｚ＝０に囲まれた部分の体積を求めよ
(3)曲面：ｚ＝ｘ^2＋ｙ^2と平面z=xに囲まれた体積を求めよ

回答No.1

info22_

(1)
逐次積分に直せば
I=∫[0,1] dx∫[x,2x](xy-x^2)^(1/2)dy
=∫[0,1] dx (2/3)[{(xy-x^2)^(3/2)}/x] [y:x,2x]
=(2/3)∫[0,1] x^2 dx
後はできますね。

(2)
x^2＋y^2=ax を変形すると
(x-(a/2))^2+y^2=(a/2)^2
円柱座標（円筒座標）変換すると
x=rcosθ,y=rsinθ
積分領域：ｘ^2＋ｙ^2≦axは
0≦r≦acosθ(-π/2≦θ≦π/2)

被積分関数；z=(x^2+y^2)/bは
z=r^2/b
dxdy=rdrdθ
となるから

求める体積Vは
V=∫[-π/2,π/2]dθ∫[0,acosθ]　(r^2/b)rdr
=(2/b)∫[0,π/2] dθ[(1/4)r^4] [r:0,acosθ]
=(2/b)∫[0,π/2] ((a^4)/4)cos^4θdθ
=((a^4)/(2b))∫[0,π/2] cos^4θdθ
=((a^4)/(2b))∫[0,π/2] (1/4)(1+cos(2θ))^2dθ
=((a^4)/(8b))∫[0,π/2] (1+cos(2θ))^2dθ
=((a^4)/(8b))∫[0,π/2] {1+2cos(2θ)+cos^2(2θ)}dθ
=((a^4)/(8b))∫[0,π/2] {1+2cos(2θ)+(1/2)+(1/2)cos(4θ)}dθ

ここからはできるだろうからやってみて下さい。

(3)
V=2∫[0,1] dx∫[0,√(x-x^2)](x^2+y^2)dy
=2∫[0,1] dx [yx^2+(1/3)y^3] [y:0,√(x-x^2)]
=2∫[0,1] {x^2(x-x^2)^(1/2)+(1/3)(x-x^2)(x-x^2)^(1/2)}dx
=(2/3)∫[0,1] (x+2x^2)(x-x^2)^(1/2)dx
x-(1/2)=tとおくとx=t+1/2,1-x=1/2-t
V=(2/3)∫[-1/2,1/2] (2t+1)(t+1)((1/2)^2-t^2)^(1/2)dt
=(2/3)∫[-1/2,1/2] (2t^2+1)((1/2)^2-t^2)^(1/2)dt
=(4/3)∫[0,1/2] (2t^2+1)((1/2)^2-t^2)^(1/2)dt
t=(1/2)sin(u)とおくと
V=(4/3)∫[0,π/2] ((1/2)sin^2(u)+1)(1/2)cos(u)(1/2)cos(u)du
=(1/3)∫[0,π/2] ((1/2)sin^2(u)+1)cos^2(u)du
=(1/3)∫[0,π/2] ((1/4)(1-cos(2u))+1)(1/2)(1+cos(2u))du
=(1/24)∫[0,π/2] (5-cos(2u))(1+cos(2u))du
=(1/24)∫[0,π/2] {5+4cos(2u)-cos^2(2u)}du
=(1/24)∫[0,π/2] {5+4cos(2u)-(1/2)(1+cos(4u))}du
この続きは自分で計算できるだろうからやってみて下さい。