求抛物面z=x^2+y^2,0<z<9的面积

2025-06-21 09:30:48
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回答1:

求抛物面z=x^2+y^2,0<z<9的面积
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