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Triple integral to find volume of ice cream cone

  1. Nov 18, 2009 #1
    1. The problem statement, all variables and given/known data
    Use a triple integral in rectangular coordinates to find the volume of the ice cream cone defined as follows

    The region R in the xy-plane is the circle of radius 1 with center at the origin.
    The sides of the cone are defined by the function z= [tex]\sqrt{x^2+y&2}[/tex]
    The top of the cone is a portion of the sphere with radius [tex]\sqrt{2}[/tex] and center at the origin.
    Repeat using triple integrals in cylindrical and spherical coordinates

    2. Relevant equations
    The problem with a picture can be found here http://www.docbenton.com/MAT240-56707/maple/MAPLE-6.pdf [Broken]


    3. The attempt at a solution

    I need some help in starting. I am not sure how to set up the triple integral. I really need the practice on this problem.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 18, 2009 #2

    LCKurtz

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    Some questions to get you started on spherical first since that is the most natural for this problem. Do you see why?

    1. What is the equation of the sphere in spherical coordinates.
    2. Can you figure out the central angle of the cone? This has to do with the limits for [itex]\phi[/itex]
    3. Do you know the spherical element of volume dV?
    4. Can you see what the limits for [itex]\rho[/itex] would be? Ditto [itex]\phi[/itex] and [itex]\theta[/itex]?

    Show us which of these you do/don't know and we can go from there.
     
  4. Nov 18, 2009 #3
    Hi. Thank you for taking the time to respond.

    1. What is the equation of the sphere in spherical coordinates? [tex]\rho[/tex]=[tex]\sqrt{2}[/tex]

    2.Can you figure out the central angle of the cone? I'm a little confused. What is the "central angle of the cone"? I believe the limits of [tex]\phi[/tex] are 0[tex]\leq[/tex][tex]\phi\leq[/tex][tex]\pi[/tex]/4

    3. Do you know the spherical element of volume dV? I do not.

    4. Can you see what the limits for rho would be? Ditto for phi and theta? The limit for theta should be 0[tex]\leq[/tex][tex]\theta[/tex][tex]\leq[/tex]2[tex]\pi[/tex]

    Sorry for the mix of latex and text. I am running out of time at the moment. I am not sure of the answer to question 3. Are you talking about the volume of sphere being 4/3[tex]\pi[/tex]r[tex]^{2}[/tex]?
     
  5. Nov 18, 2009 #4

    LCKurtz

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    Correct.
    Yes, I meant the angle from the vertical axis to the slant side. You have the [itex]\phi[/itex] limits correct.
    Well, of course you need to know dV. It should be in your book:
    [tex]dV = \rho^2\sin(\phi)d\rho d\phi d\theta[/tex]

    No, we aren't doing the volume of a sphere. Since [itex]\rho[/itex] is going to go from 0 to [itex]\sqrt 2[/itex] and you know the other limits and dV now, just put the proper limits on

    [tex]V = \int\int\int 1\ dV [/tex]

    and integrate.
     
  6. Nov 19, 2009 #5
    Wow, thank you. I have another question. Am I integrating "1" or am I integrating [tex]\rho^{2}[/tex]sin([tex]\phi[/tex]).

    Its 1 isnt it? Nevermind, I think that is obvious. I'm going to post this question just in case. Thanks again for all the help.
     
  7. Nov 19, 2009 #6

    LCKurtz

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    Remember that in spherical coordinates:

    [tex]
    dV = \rho^2\sin(\phi)d\rho d\phi d\theta
    [/tex]

    The fact that you are integrating 1 dV is what gives you the volume. If you were, for example, integrating a mass density [itex]\int\int\int_V\delta(x,y,z)\ dV[/itex] you would be calculating the total mass. The [itex]\rho^2\sin(\phi)[/itex] is built into the dV.
     
  8. Nov 21, 2009 #7
    I have worked out the problem and gotten a solution. I have the volume as 1.0017 units. Can anyone confirm this?


    My last step looked like this:

    (2/3)^(3/2) [tex]\int-cos(\pi/4)+1[/tex] d[tex]\theta[/tex]

    as theta goes from 0 to 2[tex]\pi[/tex]
     
  9. Nov 21, 2009 #8

    LCKurtz

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    That first factor should be:
    [tex]\frac {2^{\frac 3 2}}{3}[/tex]
    The rest looks right. If you were in my class I would have told you to give the answer in exact radical form, not a decimal.
     
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