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Math Help - [SOLVED] Tricky volume problem

  1. #1
    Senior Member Spec's Avatar
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    [SOLVED] Tricky volume problem

    What volume is left of a cone with the radius 2a and the height h, when you cut out a cylinder with the radius a from the cone (so that the surface of the cylinder is touching the axis of symmetry of the cone.

    The volume of the cone can be described as:
    \sqrt{x^2 + y^2} \leq z \leq h

    And the cylinder's projection on the xy-plane as:
    D= \left\{(x,y)\in R^2: (x-a)^2 + y^2 \leq a^2 \right\}

    The volume of the untouched half of the cone is already known: \frac{2}{3}\pi r^2 h

    Integrating all this is not easy though:
    \int \int_D (h - \sqrt{x^2 + y^2})dxdy

    By making the variable change u = x - a,\ v = y we get a projection on the xy-plane that can easily be turned into polar coordinates, but integrating this turned out to be quite difficult:
    \int \int_E (h - \sqrt{(u+a)^2 + v^2})dudv

    Especially in polar coordinates:
    \int \int_F (h - \sqrt{(rcos\varphi+a)^2 + r^2sin^2\varphi})r drd\varphi

    Is there an easier way to solve this problem?
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  2. #2
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    simplify your function by using some factoring
    remember cos^2(x) + sin^2(x) = 1
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  3. #3
    Senior Member Spec's Avatar
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    There is supposely an easier way to solve this problem, but I can't think of anything. Anyone?
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  4. #4
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    Quote Originally Posted by Spec View Post
    What volume is left of a cone with the radius 2a and the height h, when you cut out a cylinder with the radius a from the cone (so that the surface of the cylinder is touching the axis of symmetry of the cone.
    the equation of (the upper half of course) of the cone is z=\frac{h}{2a}\sqrt{x^2+y^2}, and the equation of the cylinder is x^2+y^2=2ax, which in polar coordinates becomes:

    r=2a \cos \theta, \ \frac{-\pi}{2} \leq \theta \leq \frac{\pi}{2}. thus: the volume of the cone cut by the cylinder = \int \int_D (h - \frac{h}{2a} \sqrt{x^2+y^2}) \ dA = h \times (\text{the area of D}) - \frac{h}{2a}\int \int_D \sqrt{x^2+y^2} \ dA=

    \pi a^2 h - \frac{h}{2a}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \int_0^{2a\cos \theta}r^2 \ dr \ d \theta = \pi a^2 h - \frac{4}{3}a^2 h\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos^3 \theta \ d \theta=\pi a^2 h - \frac{16}{9}a^2h. hence the volume left = \frac{4}{3}\pi a^2 h - \left(\pi a^2 h - \frac{16}{9}a^2h \right)=\left(\frac{\pi}{3} + \frac{16}{9} \right)a^2 h. \ \ \ \square
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  5. #5
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    Hello, Spec!

    Please check the wording of the problem.
    . . As stated, the situation is impossible.
    Also, we don't need Calculus. . .


    What volume is left of a cone with the radius 2a and the height h,
    when you cut out a cylinder with the radius a from the cone so that
    the surface of the cylinder is touching the axis of symmetry of the cone. .??

    If the radius of the cone is 2a and the radius of the cylinder is a,
    . . then the situation must look like this:
    Code:
        -         *
        :        /:\
        :       / : \
        :      /  :  \
       h:     *---+---*
        :    /|   :   |\
        :   / |   :  y| \
        :  /  |   :   |  \
        - *---*---+---*---*
            a   a   a   a

    Let the height of the cylinder by y.

    From similar right triangles, we have: . \frac{y}{a} \:=\:\frac{h}{2a} \quad\Rightarrow\quad y \:=\:\frac{h}{2}

    The volume of a cone is: . V \:=\:\frac{1}{3}\pi r^2h
    . . So we have: . V_{cone} \;=\;\frac{1}{3}\pi(2a)^2h \;=\;\frac{4}{3}\pi a^2h


    The volume of a cylinder is: . V \:=\:\pi r^2h

    . . So we have: . V_{cyl} \:=\: \pi a^2y \:=\:\pi a^2\left(\frac{h}{2}\right) \:=\:\frac{1}{2}\pi a^2h


    Therefore, the remaining volume is: . \frac{4}{3}\pi a^2h - \frac{1}{2}\pi a^2h \;=\;\boxed{\frac{5}{6}\pi a^2h}

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  6. #6
    Senior Member Spec's Avatar
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    The radius of the cone at the base is 2a. NonCommAlg's answer is correct, but I have no idea how he got there.

    If you split the base of the cone in half, and then drill a hole right through with the radius a into that half, you'll end up with the volume I was thinking of.

    I really don't get where z=\frac{h}{2a}\sqrt{x^2+y^2}, and x^2+y^2=2ax, come from.
    Last edited by Spec; August 12th 2008 at 04:31 AM.
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  7. #7
    Super Member wingless's Avatar
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    Quote Originally Posted by Soroban View Post

    If the radius of the cone is 2a and the radius of the cylinder is a,
    . . then the situation must look like this:
    Code:
        -         *
        :        /:\
        :       / : \
        :      /  :  \
       h:     *---+---*
        :    /|   :   |\
        :   / |   :  y| \
        :  /  |   :   |  \
        - *---*---+---*---*
            a   a   a   a

    Soroban, I think it should be something like this:
    Code:
    
        -         *
        :        /:\
        :       / : \
        :      /  :  \
       h:     *   +---*---+
        :    /    |    \  |
        :   /     |     \ | y
        :  /      |      \|
        - *---*---+---*---*
            a   a   a   a


    @Spec:

    z=h-\frac{h}{r}\sqrt{x^2+y^2} is the equation of a cone of radius r and height h.

    x^2+y^2 = 2ax can be derived from (x-a)^2+y^2 =a^2, which is the equation of the cylinder.
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  8. #8
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    Hi guys. Thought I'd try and illustrate what I think NonCommAlg is doing above. The integral \iint_D \left(h-\frac{h}{2a}\sqrt{x^2+y^2}\right)dA is that part of the blue cylinder inside of the the red cone: it's the height of the cone h, minus the intersecting surface of the cylinder with the cone times dA over the region D with D being the area of the base of the cylinder. Guess that's confusing though. But once you see that, then it makes sense once you figure what that volume of blue inside the cone is, then just subtract that volume from the total volume of the cone which is 4/3\pi a^2 h
    Here's the Mathematica code for the first figure in case some are interested (the code for the other two is a bit messy):
    Code:
    a = 1/2; 
    h = 1; 
    cone = RevolutionPlot3D[t/a, {t, 0, a}, 
         BoxRatios -> {1, 1, 1}, PlotStyle -> Red]
    cy = Graphics3D[{Blue, Cylinder[
             {{0, a/2, 0}, {0, a/2, h}}, a/2]}]
    Show[{cone, cy}]
    Attached Thumbnails Attached Thumbnails [SOLVED] Tricky volume problem-cylinder-cone.jpg   [SOLVED] Tricky volume problem-cylinder-cone-2.jpg   [SOLVED] Tricky volume problem-cylinder-cone-3.jpg  
    Last edited by shawsend; August 12th 2008 at 06:56 AM. Reason: correct description of volume element, added code, added two more figures
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  9. #9
    Senior Member Spec's Avatar
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    Quote Originally Posted by wingless View Post
    z=h-\frac{h}{r}\sqrt{x^2+y^2} is the equation of a cone of radius r and height h.
    Where is that equation derived from?
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  10. #10
    Super Member wingless's Avatar
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    Quote Originally Posted by Spec View Post
    Where is that equation derived from?
    It can be derived easily.

    First think of the equation z=\sqrt{x^2+y^2}. Draw it on the xz plane. Do you see the isosceles right triangle? Stretch it by h/r. Now it has the right proportion. Take the negative of it, -\frac{h}{r}\sqrt{x^2+y^2}. And add h to it in order to place the vertex at (0,0,h). There, you have it: h-\frac{h}{r}\sqrt{x^2+y^2}
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  11. #11
    Senior Member Spec's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    x^2+y^2=2ax, which in polar coordinates becomes:

    r=2a \cos \theta, \ \frac{-\pi}{2} \leq \theta \leq \frac{\pi}{2}.
    I don't understand that part.

    x = r cos\theta in polar coordinates, and the radius is \sqrt{2ax} in cartesian coordinates and should thus be \sqrt{2a \cdot rcos\theta} in polar coordinates.

    And why do you only integrate one half of the circle?
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  12. #12
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    Quote Originally Posted by Spec View Post
    I don't understand that part.

    x = r cos\theta in polar coordinates, and the radius is \sqrt{2ax} in cartesian coordinates and should thus be \sqrt{2a \cdot rcos\theta} in polar coordinates.

    And why do you only integrate one half of the circle?
    <br />
x^2+y^2=2ax ~\iff~ x^2-2ax+{\color{red}a^2} + y^2 = \color{red}a^2 ~\iff~ \boxed{(x-a)^2+y^2=a^2}

    This is the equation of the cross-section of the cylinder. It is a circle with the center C(a, 0) and the radius r = a. For each pair (x, y) you have z \in \mathbb{R}
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  13. #13
    Senior Member Spec's Avatar
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    Quote Originally Posted by earboth View Post
    <br />
x^2+y^2=2ax ~\iff~ x^2-2ax+{\color{red}a^2} + y^2 = \color{red}a^2 ~\iff~ \boxed{(x-a)^2+y^2=a^2}

    This is the equation of the cross-section of the cylinder. It is a circle with the center C(a, 0) and the radius r = a. For each pair (x, y) you have z \in \mathbb{R}
    I understand that part, but I don't understand how he got that radius in polar coordinates. My calculations show that it should be \sqrt{2ax} in carthesian coordinates and \sqrt{2a \cdot r cos\theta} in polar coordinates.
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  14. #14
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    Quote Originally Posted by Spec View Post
    I understand that part, but I don't understand how he got that radius in polar coordinates. My calculations show that it should be \sqrt{2ax} in carthesian coordinates and \sqrt{2a \cdot r cos\theta} in polar coordinates.
    1. to help you see why the equation of the cone is z=\frac{h}{2a}\sqrt{x^2+y^2}, for example look at the image of the cone in the first quadrant of yz plane, which is the line z=\frac{h}{2a}y.

    2. in polar coordinates we have x=r\cos \theta and x^2+y^2=r^2. thus x^2+y^2=2ax becomes r^2=2ar\cos \theta, which, after dividing both sides by r, gives us r=2a\cos \theta.

    the easiest way to see why \frac{-\pi}{2} \leq \theta \leq \frac{\pi}{2} is to draw the circle!
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