∬🇩 e⁻ˣ²⁻ʸ²dxdy
x=rcosθ,y=rsinθとおくと,|J|=rより
∫[0→∞]re⁻ʳ²dr・∫[0→π]dθ
＝[-(1/2)e⁻ʳ²][0→∞]・π
＝π/2
Mathematics
Undergraduate
この問題が分かりません。どなたか教えてくれませんか。
J₁₁-²²-
2²-y² dxdy,
1. T
2.
3.
4.
D = {(x, y) | y ≥ 0}
5.
T = -2-3-4-6
π
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