为自己加油

设区域D：x²+y²≤a²(a＞0)，求a的值，
使∫∫_{D2-x²-y²)dxdy=π．}

令x=rcosθ，y=rsinθ， ∫∫_D√(a²-x²-y²)dxdy= ∫₀^2πdθ∫₀^a√(a²-r²)rdr =2π∫₀^a√(a²-r²)[-(1/2)]d(a²-r²) =-(2/3)π(a²-r²)^3/2∣₀^a =2a³π/3=π, ∴a=³√(3/2)