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1. D: x^2+y^2≤2x, 即 (x-1)^2+y^2≤1, 化为极坐标为 r≤2cost, -π/2≤t≤π/2
I=∫∫<D>|xy|dσ=∫<-π/2, π/2>dt∫<0,2cost>r^2*|sintcost|rdr
=∫<-π/2, π/2>|sintcost|dt∫<0,2cost>r^3dr
=4∫<-π/2, π/2>|sintcost|(cost)^4dt
= -4∫<-π/2, 0>sint(cost)^5dt+4∫<0, π/2> sint(cost)^5dt
=[2(cost)^6/3]<-π/2, 0>-[2(cost)^6/3]<0, π/2>
=4/3.
2. 消去z,得两曲线在xOy平面内的投影为 x^2+y^2=1,
V=∫∫<D>(2-2x^2-2y^2)dσ=2∫<0, 2π>dt∫<0,1>(1-r^2)rdr
=4π∫<0,1>(r-r^3)dr=π
I=∫∫<D>|xy|dσ=∫<-π/2, π/2>dt∫<0,2cost>r^2*|sintcost|rdr
=∫<-π/2, π/2>|sintcost|dt∫<0,2cost>r^3dr
=4∫<-π/2, π/2>|sintcost|(cost)^4dt
= -4∫<-π/2, 0>sint(cost)^5dt+4∫<0, π/2> sint(cost)^5dt
=[2(cost)^6/3]<-π/2, 0>-[2(cost)^6/3]<0, π/2>
=4/3.
2. 消去z,得两曲线在xOy平面内的投影为 x^2+y^2=1,
V=∫∫<D>(2-2x^2-2y^2)dσ=2∫<0, 2π>dt∫<0,1>(1-r^2)rdr
=4π∫<0,1>(r-r^3)dr=π
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