Question

2条2重积分的题求高手解答，写一下思路，答案我有。

1.若∫∫（x^2+y^2)^1/2dxdy=4/9,其中a>0，则a=___
2.设D：x^2+y^2<=R^2,R>0且∫∫D（R^2-X^2-Y^2)^1/2dxdy=2/3π，则R=

Answer 1

1. a是什么？

2. 化成极坐标x=rcosθ, y=rsinθ, D: 0<r≤R, 0≤θ≤2π

∫∫D √(R²-x²-y²)dxdy=∫∫D r√(R²-r²)drdθ
=∫(0,2π)dθ∫(0,R)r√(R²-r²)dr

令r=Rsint, 则√(R²-r²)=Rcost, dr=Rcost dt
∫(0,R)r√(R²-r²)dr
=∫(0,π/2)R³sintcos²tdt
=-∫(0,π/2)R³cos²td(cost)
=-R³cos³t/3|(0,π/2)
=R³/3
∴∫∫D √(R²-x²-y²)dxdy
=∫(0,2π)dθ∫(0,R)r√(R²-r²)dr
=2π*R³/3
=2π/3
故R³=1
R=1

追问

1.若∫∫(x^2+y^20，则a=___

追答

x²+y²≤ax => x²+y²-ax≤0 =>(x-a/2)²+y²≤(a/2)²
积分域为以(a/2,0)为圆心, a/2为半径的圆
化成极坐标x=rcosθ, y=rsinθ, D: 0≤r≤acosθ, -π/2≤θ≤π/2
∫∫(x²+y²≤ax) √(x²+y²)dxdy=∫∫D r² drdθ
=∫(-π/2,π/2)dθ∫(0,acosθ)r²dr
=∫(-π/2,π/2) (a³cos³θ)/3 dθ
=(a³/3)∫(-π/2,π/2) cos²θ d(sinθ)
=(a³/3) (sinθ-sin³θ/3) | (-π/2,π/2)
=(a³/3) (4/3)
=4a³/9=4/9
∴a³=1
a=1

Answer 2

第一题a都不知道在哪里，不过思路是一样的。
令x=rcost，y=rsint，于是x^2+y^2=r^2，dxdy=rdrdt
代入积分就可以做了