对坐标的曲面积分 题如图所示
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对坐标的曲面积分 题如图所示
∑不为封闭曲面
所以补充平面∑1:z=1 (x^2+y^2<=1,x>=0,y>=0)(取下侧)
∑2:x=0(取前侧)
∑3:y=0(取右侧)
∑+∑1+∑2+∑3:封闭曲面,取内侧
I总=(∑+∑1+∑2+∑3)∫∫xydzdx+zdxdy
高斯公式:
=-∫∫∫[0+x+1]dxdydz
=-∫∫∫(x+1)dxdydz
使用柱坐标:
则z的积分限(r^2,1)
r的积分限(0,1)
θ的积分限(0,π/2)
=-∫(0,π/2)dθ∫(0,1)rdr∫(r^2,1)(1+r*cosθ)dz
=-∫(0,π/2)dθ∫(0,1)[(r-r^3)(1+r*cosθ)]dr
=-∫(0,π/2)dθ∫(0,1)[(r-r^3)+(r^2-r^4)*cosθ]dr
=-∫(0,π/2)[0.5*r^2-(1/4)*r^4+((1/3)r^3-(1/5)r^5)cosθ]|(0,1)dθ
=-∫(0,π/2)[0.5*-(1/4)+((1/3)-(1/5))cosθ]dθ
=-∫(0,π/2)[1/4+(2/15)*cosθ]dθ
=-[θ/4+(2/15)*sinθ](0,π/2)
=-(π/8)-(2/15)
因为:I1=(∑1)∫∫xydzdx+zdxdy
=-∫∫(Dxy)1dxdy
=-π/4
I2=(∑2)∫∫xydzdx+zdxdy=0
I3=(∑3)∫∫xydzdx+zdxdy=0
所以:I=I总-I1-I2-I3=-(π/8)-(2/15)-(-π/4)-0-0=(π/8)-(2/15)
∑不为封闭曲面
所以补充平面∑1:z=1 (x^2+y^2<=1,x>=0,y>=0)(取下侧)
∑2:x=0(取前侧)
∑3:y=0(取右侧)
∑+∑1+∑2+∑3:封闭曲面,取内侧
I总=(∑+∑1+∑2+∑3)∫∫xydzdx+zdxdy
高斯公式:
=-∫∫∫[0+x+1]dxdydz
=-∫∫∫(x+1)dxdydz
使用柱坐标:
则z的积分限(r^2,1)
r的积分限(0,1)
θ的积分限(0,π/2)
=-∫(0,π/2)dθ∫(0,1)rdr∫(r^2,1)(1+r*cosθ)dz
=-∫(0,π/2)dθ∫(0,1)[(r-r^3)(1+r*cosθ)]dr
=-∫(0,π/2)dθ∫(0,1)[(r-r^3)+(r^2-r^4)*cosθ]dr
=-∫(0,π/2)[0.5*r^2-(1/4)*r^4+((1/3)r^3-(1/5)r^5)cosθ]|(0,1)dθ
=-∫(0,π/2)[0.5*-(1/4)+((1/3)-(1/5))cosθ]dθ
=-∫(0,π/2)[1/4+(2/15)*cosθ]dθ
=-[θ/4+(2/15)*sinθ](0,π/2)
=-(π/8)-(2/15)
因为:I1=(∑1)∫∫xydzdx+zdxdy
=-∫∫(Dxy)1dxdy
=-π/4
I2=(∑2)∫∫xydzdx+zdxdy=0
I3=(∑3)∫∫xydzdx+zdxdy=0
所以:I=I总-I1-I2-I3=-(π/8)-(2/15)-(-π/4)-0-0=(π/8)-(2/15)
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