Urgent! Spherical coordinates?

iv23361512

2012-02-27 4:57 am

Evaluate the integral of f(x,y,z)=z(x^2+y^2+z^2)^(-3/2), over the part of the ball x^2+y^2+z^2<=36 and z>=3.

I have tried to do it, but got the wrong answer, which is:
∫(θ in [0, 2π]) ∫(φ in [π/3, 0]) ∫(ρ in [3/cosφ, 0]) (ρ cos φ) (ρ^-3) * (ρ^2 sin φ dρ dφ dθ)=3π
Help me, please!

更新1:

Thanks for the really quick reply, it was really really helpful!! But for the answer, shouldn't it be (3/2)pi from the calculations?

回答 (1)

2012-02-27 5:15 am

✔ 最佳答案

Note that
x^2 + y^2 + z^2 = 36 ==> ρ = 6
and z = 3 ==> ρ cos φ = 3 ==> ρ = 3/cos φ.

These intersect when 3/cos φ = 6 ==> φ = π/3.

So, ∫∫∫ z(x^2+y^2+z^2)^(-3/2) dV
= ∫(θ = 0 to 2π) ∫(φ = 0 to π/3) ∫(ρ = 3/cos φ to 6) (ρ cos φ) ρ^(-3) * (ρ^2 sin φ dρ dφ dθ)
= 2π ∫(φ = 0 to π/3) ∫(ρ = 3/cos φ to 6) cos φ sin φ dρ dφ
= 2π ∫(φ = 0 to π/3) (6 - 3/cos φ) cos φ sin φ dφ
= 2π ∫(φ = 0 to π/3) (-6 cos φ + 3) * -sin φ dφ
= 2π (-3 cos^2(φ) + 3 cos φ) {for φ = 0 to π/3}
= 3π.

I hope this helps!

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