質問<3754>2009/7/15
∬xdxdy {(x、y):x^2+y^2≦1、x≧0、y≧0} ★希望★完全解答★
お便り2009/7/29
from=phaos
[解 1] 積分領域が四分円なので, 極座標変換します。 即ち x = r cosθ, y = r sinθ, 0 < r ≦ 1, 0 ≦ θ ≦ π/2. Jacobian は ∂(x, y)/ ∂(r, θ) = r なので, 与式 = ∫_(0 < r ≦ 1, 0 ≦ θ ≦ π/2) r^2 cosθdr dθ = [r^3/3]_0^1[sinθ]_0^(π/2) = 1/3. [解 2] 与式 = ∫_(y = 0)^1∫_(x = 0)^(√(1 - y^2)) x dx dy = ∫_(y = 0)^1 [x^2/2]_(x = 0)^(√(1 - y^2)) dy = (1/2) ∫_(y = 0)^1 (1 - y^2) dy = (1/2) [y - y^3/3]_0^1 = (1/2)(2/3) = 1/3.