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Math Help - Volume of Cone in R3

  1. #1
    Ant
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    Volume of Cone in R3

    I'm trying to compute the area of the Cone (Z = sqrt[ x^2 + y^2] below the plane z = 2.


    Using cylindricl polars coordinates:

    phi: [ -pi/2, pi/2]
    r: [ 0, 2]
    z: [ r, 2]

    with the Jacobian being r.

    I get 4/3 pi which I think is the correct answer.

    However, using Spherical polar coordinates and using the following limits:

    Phi: [ -pi/2, pi/2 ]
    Theta: [ 0, pi/4 ]
    r: [ 0, 2 ]

    with the Jacobian being r^2 sin (theta) I'm getting a different answer. Could anyone tell me if my limits are correct?

    Thanks!
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  2. #2
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    Quote Originally Posted by Ant View Post
    I'm trying to compute the area of the Cone (Z = sqrt[ x^2 + y^2] below the plane z = 2.


    Using cylindricl polars coordinates:

    phi: [ -pi/2, pi/2]
    r: [ 0, 2]
    z: [ r, 2]

    with the Jacobian being r.

    I get 4/3 pi which I think is the correct answer.

    However, using Spherical polar coordinates and using the following limits:

    Phi: [ -pi/2, pi/2 ]
    Theta: [ 0, pi/4 ]
    r: [ 0, 2 ]

    with the Jacobian being r^2 sin (theta) I'm getting a different answer. Could anyone tell me if my limits are correct?

    Thanks!
    \int_{0}^{2\pi} \int_{0}^{2} \int _{r}^{2}rdzdrd\phi

    \int_{0}^{2\pi} \int_{0}^{2} r(2-r)drd\phi

    \int_{0}^{2\pi}  r^2-\frac{1}{3}r^3\bigg|_{0}^{2}d\phi

    \frac{4}{3}\int_{0}^{2\pi}  d\phi=\frac{8\pi}{3}

    Remember that geometry that the volume formula is

    V=\frac{1}{3}\pi r^2 h=\frac{1}{3}\pi (2)^2(2)=\frac{8\pi}{3}

    for the 2nd one r goes from

    r=0..\frac{2}{\cos(\theta)}

    why?
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  3. #3
    Ant
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    Sorry! I've missed a vital piece of information. It is given that X > 0!

    That's why my limits run from -pi/2 to pi/2.
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    Quote Originally Posted by Ant View Post
    Sorry! I've missed a vital piece of information. It is given that X > 0!

    That's why my limits run from -pi/2 to pi/2.
    This will just change phi as you have them above

    \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{2} \int _{r}^{2}rdzdrd\phi

    \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{2} r(2-r)drd\phi

    \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}  r^2-\frac{1}{3}r^3\bigg|_{0}^{2}d\phi

    \frac{4}{3}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}  d\phi=\frac{4\pi}{3}

    Remember that geometry that the volume formula is

    V=\frac{1}{3}\pi r^2 h=\frac{1}{3}\pi (2)^2(2)=\frac{8\pi}{3}

    for the 2nd one

    \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{2\pi}\int_  {0}^{\frac{2}{\cos(\theta)}}r^2\sin(\theta)dr d\theta d\phi

    why?
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  5. #5
    Ant
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    Was the area that I computed in the second case a portion of a sphere? So it was curved at the top were z=2 (if that makes sense) whereas I'm trying to find the area of a cone so it's a straight line.
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    Quote Originally Posted by Ant View Post
    Was the area that I computed in the second case a portion of a sphere? So it was curved at the top were z=2 (if that makes sense) whereas I'm trying to find the area of a cone so it's a straight line.
    Yes it looks kind of like an ice cream cone.
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  7. #7
    Ant
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    I think I know where the limits comes from:

    Unlike for a sphere the r is changes, as we track round with theta changing form zero to pi/4 the distance form the centre isn't constant.

    Here r it is given by the hypotenuse of a triangle.

    Cos (theta) = adjacent/hyp = z/r Where z =2, so we have;

    r = 2/cos(theta)
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  8. #8
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    Quote Originally Posted by Ant View Post
    I think I know where the limits comes from:

    Unlike for a sphere the r is changes, as we track round with theta changing form zero to pi/4 the distance form the centre isn't constant.

    Here r it is given by the hypotenuse of a triangle.

    Cos (theta) = adjacent/hyp = z/r Where z =2, so we have;

    r = 2/cos(theta)
    Yes that is correct.
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