求这两个二重积分
1个回答
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第一题,化为极坐标
x^2+y^2=π^2,p=π
x^2+y^2=4π^2,p=2π
a∈[0,2π]
∫∫sin√(x^2+y^2)dσ
=∫[π,2π]∫[0,2π]sinp*pdpda
=∫[π,2π]sinp*pdp∫[0,2π]da
=2π∫[π,2π]sinp*pdp
=2π∫[π,2π]sinp*pdp
=-2π∫[π,2π]pdcosp
=-2π*pcosp[π,2π]+2π∫[π,2π]cospdp
=2π*cosp[π,2π]
=4π
∫∫(x^2+y^2)dσ
=∫[a,3a]∫[y-a,y](x^2+y^2)dxdy
=∫[a,3a](x^3/3+xy^2)[y-a,y]dy
=∫[a,3a](2ay^2-a^2y+a^3/3)dy
=(2ay^3/3-a^2y^2/2+a^3y/3)[a,3a]
=18a^4-9/2a^4+a^4-2/3a^4+a^4/2-a^4/3
=14a^4
x^2+y^2=π^2,p=π
x^2+y^2=4π^2,p=2π
a∈[0,2π]
∫∫sin√(x^2+y^2)dσ
=∫[π,2π]∫[0,2π]sinp*pdpda
=∫[π,2π]sinp*pdp∫[0,2π]da
=2π∫[π,2π]sinp*pdp
=2π∫[π,2π]sinp*pdp
=-2π∫[π,2π]pdcosp
=-2π*pcosp[π,2π]+2π∫[π,2π]cospdp
=2π*cosp[π,2π]
=4π
∫∫(x^2+y^2)dσ
=∫[a,3a]∫[y-a,y](x^2+y^2)dxdy
=∫[a,3a](x^3/3+xy^2)[y-a,y]dy
=∫[a,3a](2ay^2-a^2y+a^3/3)dy
=(2ay^3/3-a^2y^2/2+a^3y/3)[a,3a]
=18a^4-9/2a^4+a^4-2/3a^4+a^4/2-a^4/3
=14a^4
追问
你看,第一题的最后三步有点问题是不
追答
第一题,化为极坐标
x^2+y^2=π^2,p=π
x^2+y^2=4π^2,p=2π
a∈[0,2π]
∫∫sin√(x^2+y^2)dσ
=∫[π,2π]∫[0,2π]sinp*pdpda
=∫[π,2π]sinp*pdp∫[0,2π]da
=2π∫[π,2π]sinp*pdp
=-2π∫[π,2π]pdcosp
=-2π*pcosp[π,2π]+2π∫[π,2π]cospdp
=-2π*pcosp[π,2π]
=-4π^2-2π^2
=-6π^2
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