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Find the volume of a cone using spherical coordinates

  1. Feb 19, 2009 #1
    Find the volume of the portion of cone z^2 = x^2 + y^2 bounded by the planes z = 1 and z = 2 using spherical coordinates

    I am having trouble coming up with the limits






    Relevant equations
    dV = r^2*sin(theta)*dr*d(theta)*d(phi)
    r = sqrt(x^2+y^2+z^2)

    the problem is actually 2 parts, the 2nd part asks to evaluate by cylindrical coordinates and I obtain 7pi/3 which i know is right, I just cant come up with the limits

    Attempt
    for r i have from sqrt2 to 2sqrt2
    for theta i have from 0 to pi/4
    and for phi i have from 0 to 2pi
     
  2. jcsd
  3. Feb 19, 2009 #2

    HallsofIvy

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    In spherical coordinates, [itex]x= \rho cos(\theta)sin(\phi)[/itex], [itex]y= \rho sin(\theta)sin(\phi)[/itex], and [itex]z= \rho cos(\phi)[/itex]
    Putting those into [itex]z^2= x^2+ y^2[/itex] becomes [itex]\rho^2 cos^2(\phi)= \rho^2 cos^2(\theta)sin^2(\phi)+ \rho^2 cos^2(\theta)sin^2(\phi)[/itex][itex]\rho^2 sin^2(\phi)[/itex] which reduces to [itex]cos^2(\phi)= sin^2(\phi)[/itex]. For [itex]\phi[/itex] between 0 and [itex]\pi/2[/itex], that is true only for [itex]\phi= \pi/4[/itex].

    Now, for [itex]z= \rho cos(\phi)= 1[/itex] then [itex]\rho= 1/cos(\phi)[/itex] for [itex]z= \rho cos(\phi)= 2[/itex] then [itex]\rho= 2/cos(\phi)[/itex] the limits of integration are: [itex]rho[/itex] from [itex]1/cos(\phi)[/itex] to [itex]2/cos(\phi)[/itex], [itex]\phi[/itex] from 0 to [itex]\pi/4[/itex], and [itex]\theta[/itex] from 0 to [itex]2\pi[/itex].

    It looks to me like cylindrical coordinates should be much easier! Again, [itex]\theta[/itex] runs from 0 to [itex]2\pi[/itex], z obviously runs from 1 to 2, and, since [itex]z^2= x^2+ y^2= r^2[/itex], z= r (z is positive) so r runs from 0 to z.
     
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