举报
sybase0217
上半球方程为:z=√(a²-x²-y²) ∂z/∂x=-x/√(a²-x²-y²),∂z/∂y=-y/√(a²-x²-y²) 因此:1+(∂z/∂x)²+(∂z/∂y)²=1+x²/(a²-x²-y²)+y²/(a²-x²-y²)=a²/(a²-x²-y²) 则:dS=√[1+(∂z/∂x)²+(∂z/∂y)²]dxdy=[a/√(a²-x²-y²)]dxdy ∫∫√(a²+x²+y²) dS =∫∫√(a²+x²+y²)*[a/√(a²-x²-y²)]dxdy =a∫∫1dxdy 被积函数为1,积分结果为区域面积πa²/2 =πa³/2