面积=∫∫D √1+4x²+4y² dxdy
=∫∫D √1+4p² pdpdθ
=∫(0,2π)dθ∫(0,√2)√1+4p² pdp
=π/4∫(0,√2)√1+4p² d(1+4p²)
=π/4 · 2/3 (1+4p²)的2分之3次方 (0,√2)
=π/6 [27-1]
=13π/3
面积=∫∫D √1+4x²+4y² dxdy
=∫∫D √1+4p² pdpdθ
=∫(0,2π)dθ∫(0,√2)√1+4p² pdp
=π/4∫(0,√2)√1+4p² d(1+4p²)
=π/4 · 2/3 (1+4p²)的2分之3次方 (0,√2)
=π/6 [27-1]
=13π/3