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Double integral polar coordinates trouble please help!

  1. Jan 31, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider the volume of a solid bounded by the cone: z = sqrt(x^2 + y^2) and the top half of the sphere x^2 + y^2 + z^2 = 18 that is for z >= 0

    Using cylindrical coordinates, express the volume as a double integral.


    2. Relevant equations
    easy to sketch.. we can pretty easily figure r = 3 and theta = 2 pi


    3. The attempt at a solution

    I got integral of sqrt(18 - r^2) r.dr.d(theta) over r = { (rcos(theta), rsin(theta) | 0 <= r <= 3, 0 <= theta <= 2pi


    but answer says... [sqrt(18-r^2) - r ] r.dr.d(theta) where have i gone wrong


    thanks in advance for the help!
     
  2. jcsd
  3. Jan 31, 2009 #2

    Hootenanny

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    Remember that we are dealing with cylindrical coordinates here, *not* spherical coordinates. So r2 = x2 + y2 and the volume element is r.dr.d[itex]\theta[/itex] dz.

    So let's try again. Can you express the equation of the cone in terms of *cylindrical* coordinates?
     
  4. Jan 31, 2009 #3
    im so lost :( where did the other r come from?
     
  5. Jan 31, 2009 #4
    also r.dr.d(theta).dz is triple integral not double?
     
  6. Jan 31, 2009 #5

    Hootenanny

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    Which r? The expression, [itex]r dr d\theta dz[/itex] is simply the volume element in cylindrical coordinates. In Cartesian coordinates, the equivalent expression would be [itex] dx dy dz[/itex].
    Indeed it is. Volume integrals *must* be triple integrals.
    However, triple integrals may be expressed as a double integral by evaluating one of the integrals, in this case the integral with respect to z.
     
  7. Jan 31, 2009 #6
    [sqrt(18-r^2) - r ] i mean the r bold and underlined.

    So i went for z limits o to sqrt(18)

    with R = {(x,y,z) in polar form | 0 <= z <= sqrt(18), 0 <=r <= 3, 0 <= theta <= 2pi}

    for what integral??? this is so confusing.. its just a Q for a practise exam is there anyway you could work it through abit? I find i understand much better if i can look at it to see how it works.. thankyou
     
  8. Jan 31, 2009 #7

    Hootenanny

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    We'll get to that eventually.
    That's not the way it works here. I'll guide you through the question, which will help you understand it better, but I won't provide a complete or partial solution.

    Can you try answering my question:
    In other words instead of writing

    [tex]z = f\left(x,y\right)[/tex]

    Write it in the form

    [tex] z = g\left(r,\theta\right)[/tex]
     
  9. Jan 31, 2009 #8
    so z=r for the cone in cylindrical form. now we are minusing the area from under the cone from the area under the half sphere right? so thats where we get our sqrt(18-r^2) - r from? because the z value ranges from z= sqrt(18-r^2) to z=r
     
  10. Jan 31, 2009 #9

    Hootenanny

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    Correct.
    Let's take it one step at a time. So, as you correct write we also have the equation of the upper half sphere:

    [tex]z = \sqrt{18-r^2}[/tex]

    and for the cone:

    [tex]z = r[/tex]

    Now, we need to figure out the limits. Note that the limits will be different for the sphere and cone. What are the limits for the sphere?
     
  11. Jan 31, 2009 #10
    Sorry had to sleep.. limits for sphere would be sqrt(18-r^2) and 0 and for the cone would be sqrt(18) and 0?
     
  12. Feb 1, 2009 #11

    Hootenanny

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    You need to be careful here, either the upper or lower limit in each case should be a function of the radius. Your limits for there sphere are correct: the smallest value z can take is zero (at the origin) and the largest value z can take is on the surface of the sphere (i.e. z = (18-r2)1/2.

    Now can you apply the same logic to the cone? What is the smallest value that z can take inside the cone. What is the largest value?
     
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