Question

一道高数题，急求

如图，麻烦给出步骤，谢谢了

Answer 1

简单计算一下即可，答案如图所示

Answer 2

积分域 D：x^2+(y-1/2)^2 = 1/4 右半圆， 面积是 S = (1/2)π(1/2)^2 = π/8.
二重积分是常数，不妨设 ∫∫<D> f(u,v)dudv = C,
f(x, y) = √(1-x^2-y^2) - (8/π)C, 两边在 D 上作二重积分， 得
∫∫<D> f(x, y)dxdy = ∫∫<D> √(1-x^2-y^2)dxdy - (8/π)CS
二重积分与积分变量无关， 左边就是 C， 又 S = π/8， 则
C = ∫∫<D> √(1-x^2-y^2)dxdy - C， 解得
C = (1/2) ∫∫<D>√(1-x^2-y^2)dxdy
= (1/2) ∫<0, π/2>dt∫<0, sint>√(1-r^2)rdr
= (-1/4) ∫<0, π/2>dt∫<0, sint>√(1-r^2)d(1-r^2)
= (-1/4)(2/3) ∫<0, π/2>dt[(1-r^2)^(3/2)]<0, sint>
= (1/6)∫<0, π/2>[1-(cost)^3]dt
= (1/6) {π/2 - ∫<0, π/2>[1-(sint)^2]dsint}
= (1/6) {π/2 - [sint - (sint)^3/3]<0, π/2>}
= (1/6) (π/2 - 2/3),
f(x, y) = √(1-x^2-y^2) - (8/π)(1/6) (π/2 - 2/3)
= √(1-x^2-y^2) - 2/3 + 8/(9π)