重積分

質問

次の値をもとめてください
(1)∬D x^2dxdy,D:(x-1)^2+y^2<=1,0<=y
(2)∬D logx/y^2dxdy,D:1<=x<=2,1<=y<=x
(3)∬D (x^2+y^2)dxdy,D:x+y<=1,0<=x,y
(4)∬D √4x^2-y^2dxdy,D:0<=y<=x<=1
(5)∬D dxdy/(1+x^2+y^2)^2,D:(x^2+y^2)^2<=x^2-y^2

途中計算も教えてください。
すいませんが至急おねがいします。

回答No.1

muturajcp · ベストアンサー

(1)
x=1+cost,(π≧t≧0)
∬_{(x-1)^2+y^2≦1,0≦y}(x^2)dxdy
=∫_{0～2}{x^2∫_{0～√(2x-x^2)}dydx
=∫_{0～2}{x^2√(2x-x^2)}dx
=∫_{π～0}[{(1+cost)^2}sint](-sint)dt
=∫_{0～π}{1+2cost+(cost)^2}(sint)^2dt
=∫_{0～π}{(sint)^2+2cost(sint)^2+((sin2t)/2)^2}dt
=∫_{0～π}{(1-cos2t)/2+cost-cos3t+(1-cos4t)/8}dt
=[t/2-(sin2t)/4+sint-(sin3t)/3+t/8-(sin4t)/32]_{0～π}
=(π/2)+(π/8)
=5π/8

(2)
x=e^t,(0≦t≦log2)
∬_{1≦x≦2,1≦y≦x}(logx)/y^2dxdy
=∫_{1～2}(logx)(∫_{1～x}y^{-2}dy)dx
=∫_{1～2}(logx)([-y^{-1}]_{1～x})dx
=∫_{1～2}(logx)([1-x^{-1})dx
=∫_{0～log2}(te^t-t)dt
=∫_{0～log2}(te^t)dt-∫_{0～log2}tdt
=[te^t]_{0～log2}-∫_{0～log2}(e^t)dt-[(t^2)/2]_{0～log2}
=2(log2)-1-{(log2)^2}/2

(3)
∬{x+y≦1,0≦x,y}(x^2+y^2)dxdy
=∫_{0～1}∫_{0～1-y}(x^2+y^2)dxdy
=∫_{0～1}([(x^3)/3+xy^2]_{0～1-y})dy
=∫_{0～1}[{(1-y)^3}/3+(1-y)y^2]dy
=∫_{0～1}[{(1-y)^3}/3+(1-y)y^2]dy
=∫_{0～1}[(1-3y+6y^2-4y^3)/3]dy
=[(y/3)-{(y^2)/6}+{(2y^3)/3}-{(y^4)/3}]_{0～1}
=(1/3)-(1/6)+(2/3)-(1/3)
=1/2

(4)
y=2xsint,(0≦t≦π/6)
∬_{0≦y≦x≦1}√(4x^2-y^2)dxdy
=∫_{0～1}∫_{0～π/6}2x^2(1+cos2t)dtdx
=∫_{0～1}x^2[2t+sin2t]_{0～π/6}dx
=(π/3+√3/2)∫_{0～1}x^2dx
=(π/9)+{(√3)/6}

(5)
x=rcost,y=rsint,z=1+r^2
∬_{(x^2+y^2)^2≦x^2-y^2}{1/(1+x^2+y^2)^2}dxdy
=∬_{r^2≦cos2t}{r/(1+r^2)^2}drdt
=∫_{-π/4≦t≦π/4,3π/4≦t≦5π/4}∫_{1～1+cos2t}{1/(2z^2)}dzdt
=∫_{-π/4≦t≦π/4,3π/4≦t≦5π/4}[-1/(2z)]_{1～1+cos2t}dt
=∫_{-π/4≦t≦π/4,3π/4≦t≦5π/4}[cos2t/{2(1+cos2t)}]dt
=∫_{-π/4≦t≦π/4,3π/4≦t≦5π/4}{(1/2)-[1/{4(cost)^2}]}dt
=[(t/2)-(1/4)tant]_{-π/4≦t≦π/4,3π/4≦t≦5π/4}
=π/2

お礼

ありがとうございます。
助かります。