Question

请问这个高数怎么求？

Answer 1

令：
x = rcosθ
y = rsinθ
则：0≤θ≤2π , 0≤r≤√(a^2-r^2) ; dxdy = rdrdθ

∵ 0<b≤z≤a
∴ z = √(a^2-x^2-y^2)

z'x = -x/√(a^2-x^2-y^2) = -x/z
z'y = -y/√(a^2-x^2-y^2) = -y/z

√[1+(z'x)^2 + (z'y)^2] = √[(z^2+x^2+y^2)/z^2] = a/z = a/√(a^2-x^2-y^2) =a/√(a^2-r^2)

∫∫√[1+(z'x)^2 + (z'y)^2] dxdy
=∫[0,2π]dθ ∫[0,√(a^2-b^2)] a/√(a^2-r^2) rdr
= 2π * a*∫[0,√(a^2-b^2)] 1/√(a^2-r^2) rdr
= aπ *∫[0,√(a^2-b^2)] 1/√(a^2-r^2) dr^2
= aπ *{ -2√(a^2-r^2) |[0,√(a^2-b^2)] }
= 2aπ *(a-b)