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Changing order of integration in spherical coordinates 
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#1
Jun2710, 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! 


#2
Jun2710, 05:42 AM

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PF Gold
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Your answer is correct. The solution is wrong.



#3
Jun2710, 07:30 AM

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Many thanks!



#4
Jun2710, 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. 


#5
Jun2710, 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{4r^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. 


#6
Jun2710, 06:24 PM

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