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Math Help - Volume of a Cone

  1. #1
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    Question Volume of a Cone

    Can someone help me with calculating the volume of a Cone by taking a vertical cross-sectional area of the cone and integrating it? I know how to solve it for the horizontal cross-section, which gives an area of circle. But if we take a vertical cross-section then we get a Hyperbola, am I correct?

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  2. #2
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    Quote Originally Posted by bilalsaeedkhan View Post
    Can someone help me with calculating the volume of a Cone by taking a vertical cross-sectional area of the cone and integrating it? I know how to solve it for the horizontal cross-section, which gives an area of circle. But if we take a vertical cross-section then we get a Hyperbola, am I correct? <=== No

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    If the cut is parallel to the axis of the cone then you'll get parabolas as cross-sections.

    Have a look here: Conic section - Wikipedia, the free encyclopedia
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    Quote Originally Posted by earboth View Post
    If the cut is parallel to the axis of the cone then you'll get parabolas as cross-sections.

    Have a look here: Conic section - Wikipedia, the free encyclopedia
    My case is the third figure in your link, which is a hyperbola. Is the cross-section that I have taken in my original attachment a parabola or a hyperbola?
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  4. #4
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    It is a lot easier to do it the other way but here it goes.

    The equation of a cone is x^2+y^2=R^2z^2.

    Suppose the radius of the base at z=h is r. Then we have Rh=r or R=\dfrac{r}{h}

    Now we let z\in\left[0,h\right]

    Next we have the equation of a general plane Ax+By+Cz=D. We want to construct a plane \mathcal{P}||(\hat{\textbf{j}},\hat{\textbf{k}})

    This would mean that x=const or \mathcal{P}\rightarrow x=x_0.

    Now lets intersect this plane with our cone. This gives us the curve

    R^2z^2-y^2=x_0^2 or y=\pm\sqrt{R^2z^2-x_0^2}

    Now at z=h y=\pm\sqrt{r^2-x_0^2} which will be our limit of integration. Keep in mind that the vertex of the cone is at z=0. We know that z>0 therefore z=\dfrac{\sqrt{y^2+x_0^2}}{R}

    Next we find the area A(x_0)=2hr-\int_{-\sqrt{r^2-x_0^2}}^{\sqrt{r^2-x_0^2}}\dfrac{\sqrt{y^2+x_0^2}}{R}\,dy= and I'm going to stop here as it is getting quite messy. I guess that helps to a point but I don't see any point in continuing.
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    Thank you! It is complicated beyond my scope but thanks nonetheless.
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  6. #6
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    Quote Originally Posted by bilalsaeedkhan View Post
    My case is the third figure in your link, which is a hyperbola. Is the cross-section that I have taken in my original attachment a parabola or a hyperbola?
    Of course you are right and I apologize for the confusion.

    Obviously a hyperbolic cross-section is hard to handle as fobos3 has demonstrated. So I suggest to use parabolic cross-sections.

    According to Archimedes a segment of a parabola has an area

    A = \frac43 \cdot \underbrace{\frac12 \cdot 2 b \cdot h}_{area\ of\ triangle} = \frac43 b h

    Use the proportion \dfrac hs = \dfrac{2R-x}{2R}~\implies~h=\dfrac{s(2R-x)}{2R}

    Use Pthagorian theorem to calculate the length of b = \sqrt{R^2-(R-x)^2}

    Caution: This step probably contains a mistake which causes a wrong final result! I wasn't able to spot my error.

    Now the colume of the cone should be:

    V=\displaystyle{\int}_0^{2R}\left(\frac43 \cdot \dfrac{s(2R-x)}{2R} \cdot \sqrt{R^2-(R-x)^2}\right) dx

    My computer tells me that V = \dfrac \pi3 \cdot R^2 \cdot s which is definitely too large.

    (... but it looks like a very near miss)
    Attached Thumbnails Attached Thumbnails Volume of a Cone-kegelvol_integral.png  
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  7. #7
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    Thanks, I'll try it again using your setup. hopefully i'll get something.
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