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Changing order of integration in spherical coordinates

by beowulf.geata
Tags: coordinates, integration, order, spherical
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beowulf.geata
#1
Jun27-10, 01:55 AM
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1. The problem statement, all variables and given/known data

Let D be the region bounded below by the plane z=0, above by the sphere x^2+y^2+z^2=4, and on the sides by the cylinder x^2+y^2=1. Set up the triple integral in spherical coordinates that gives the volume of D using the order of integration dφdρdθ.


2. Relevant equations

The solution says that D is:



3. The attempt at a solution

I thought that the solution was:



Could you please tell me where I’m going wrong? Many thanks!
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vela
#2
Jun27-10, 05:42 AM
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Your answer is correct. The solution is wrong.
beowulf.geata
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Jun27-10, 07:30 AM
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Many thanks!

LCKurtz
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Jun27-10, 04:18 PM
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Changing order of integration in spherical coordinates

I'm afraid I don't agree with vela. The answer given is correct. The problem is the the value of φ on the cylinder depends on ρ. The equation of the cylinder is r = 1 where r is the polar radius. This gives r = ρsin(φ) = 1. If you are at the top of the cylinder φ = π/6 and ρ = 2 while at the bottom of the cylinder φ = π/2 and ρ = 1. φ is the function of ρ given by φ = sin-1(1/ρ). In the second integral, if you start at z axis and move in the φ direction, you start at φ = π/6 and how far you move depends on the value of ρ, and that value is sin-1(1/ρ) That gives the inner integral limits. Then ρ goes from 1 to 2 etc.

[Edit] Nope, I take it back. vela is right as he explains below.
vela
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Jun27-10, 05:19 PM
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You're describing the volume bounded by the cylinder and the sphere ρ=1, but the volume of D also include the region between ρ=0 and ρ=1 where φ can run unrestricted from 0 to π/2. That's what the OP's third integral corresponds to.

It's easy to write the volume in terms of cylindrical coordinates:

[tex]V=\int_0^1\int_0^{2\pi}\int_0^{\sqrt{4-r^2}} r\,dz\,d\theta\,dr[/tex]

This integral evaluates to the same result as the OP's answer, but not the answer from the solution.
LCKurtz
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Jun27-10, 06:24 PM
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Quote Quote by vela View Post
You're describing the volume bounded by the cylinder and the sphere ρ=1, but the volume of D also include the region between ρ=0 and ρ=1 where φ can run unrestricted from 0 to π/2. That's what the OP's third integral corresponds to.

It's easy to write the volume in terms of cylindrical coordinates:

[tex]V=\int_0^1\int_0^{2\pi}\int_0^{\sqrt{4-r^2}} r\,dz\,d\theta\,dr[/tex]

This integral evaluates to the same result as the OP's answer, but not the answer from the solution.
Ahhh yes, you're right. That's easy to miss; puts me in good company with the author And just emphasizes the point that you should choose your coordinate system and order of integration carefully, trying not to make the problem trickier than it is.


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