用极坐标计算二重积分 ∫∫√(1-x^2-y^)/(1+x^2+y^2)dxdyD:x^2+y^2≤1,x≥0,y≥0
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/05 10:47:37
用极坐标计算二重积分
∫∫√(1-x^2-y^)/(1+x^2+y^2)dxdy
D:x^2+y^2≤1,x≥0,y≥0
∫∫√(1-x^2-y^)/(1+x^2+y^2)dxdy
D:x^2+y^2≤1,x≥0,y≥0
极坐标下D:x^2+y^2≤1,x≥0,y≥0可表示为
0≤r≤1,0≤θ≤π/2
∫∫√(1-x^2-y^2)/(1+x^2+y^2)dxdy
=∫(0,π/2)dθ∫(0,1)[(1-r^2)/(1+r^2)]rdr
=π/2∫(0,1)[(1-r^2)/(1+r^2)]rdr,
=π/2∫(0,1)r/(1+r^2)dr-(π/2)∫(0,1)r^3/(1+r^2)dr,
=π/2∫(0,1)r/(1+r^2)dr-(π/2)∫(0,1)[(r^3+r)-r]/(1+r^2)dr
=2*(π/2)∫(0,1)r/(1+r^2)dr-(π/2)∫(0,1)rdr
=(π/2)ln(1+r^2)|(0,1)-(π/2)*(1/2)r^2|(0,1)
=(π/2)*(ln2-1/2)
再问: 只有D求对了 后面就太离谱了
再答: 如图:
0≤r≤1,0≤θ≤π/2
∫∫√(1-x^2-y^2)/(1+x^2+y^2)dxdy
=∫(0,π/2)dθ∫(0,1)[(1-r^2)/(1+r^2)]rdr
=π/2∫(0,1)[(1-r^2)/(1+r^2)]rdr,
=π/2∫(0,1)r/(1+r^2)dr-(π/2)∫(0,1)r^3/(1+r^2)dr,
=π/2∫(0,1)r/(1+r^2)dr-(π/2)∫(0,1)[(r^3+r)-r]/(1+r^2)dr
=2*(π/2)∫(0,1)r/(1+r^2)dr-(π/2)∫(0,1)rdr
=(π/2)ln(1+r^2)|(0,1)-(π/2)*(1/2)r^2|(0,1)
=(π/2)*(ln2-1/2)
再问: 只有D求对了 后面就太离谱了
再答: 如图:
用极坐标计算二重积分 ∫∫√(1-x^2-y^)/(1+x^2+y^2)dxdyD:x^2+y^2≤1,x≥0,y≥0
利用极坐标计算二重积分 ∫∫(x+y)/(x^2+y^2)dxdy,其中D为x^2+y^2=1
使用极坐标计算二重积分∫∫(4-x^2-y^2)^(1/2)dxdy ,D的区域为x^2+y^2=0所围.
二重积分化极坐标计算∫∫X^2+Y^2dxdy区间 0
计算二重积分∫∫|y-x^2|dxdy,其中区域D={(x,y)|-1
计算二重积分I=∫∫(1+X+2y)dxdy ,D={(x,y) | 0≤x≤2,-1≤y≤3}
∫∫√1-x^2-y^2/1+x^2+y^2dxdy,其中D为区域x^2+y^2≤1的二重积分计算
计算二重积分 ∫ ∫D e^(x^2+y^2) dxdy,其中 D:x^2+y^2≤1
∫(0,1)dx∫(x^2,x)(x^2+y^2)^0.5求二重积分
用二重积分计算体积 x+y+z=3 x^2+y^2=1 z=0
∫∫(y/x)^2dxdy,D为曲线y=1/x,y=x,y=2所围成的区域计算二重积分
计算二重积分,∫∫(x+y)dxdy,其中D为x^2+y^2≤x+y