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Thread: Inverted Cone

  1. #1
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    Inverted Cone

    We want to use an inverted cone with a radius of 9cm to dispense 100mL of a chemical. To be certain that we dispense the correct quantity we want markings in 100mL increments on the outside of the cone. Where should we place the markings?(accurate to .1 of a mm).
    The volume of the inverted cone can be found with V=(32a^3)/27 where a is the radius of the cone

    So far all I have done is sub in the radius to work out the volume

    $\displaystyle V=(32*9^3)/27$
    $\displaystyle V=864$

    Any help would be greatly appreciated
    Last edited by FLOTUS; Aug 13th 2010 at 09:23 PM.
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  2. #2
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    What exactly is the shape of the cone? That is, what is the total height when the radius is 9 cm? The volume of a cone, of radius a and height h, is $\displaystyle \frac{\pi}{3}a^2h$. You say that the volume is given by $\displaystyle \frac{32a^3}{27}$ which leads me to the conclusion that the height is $\displaystyle \frac{a}{9\pi}$. With a= 9 cm, that gives $\displaystyle h= \frac{9}{9\pi}= \frac{1}{\pi}$ or about 0.318 cm. I find that hard to believe!

    If a cone has radius R and height H, then for any given height, h, above the vertex, the radius at that height is given by $\displaystyle \frac{r}{h}= \frac{R}{H}$ or $\displaystyle r= \frac{R}{H}r$. A circle with that radius has area $\displaystyle \pi r^2= \pi\frac{R^2}{H^2}h^2$ and, if we think of it as a thin disk of height dh, volume $\displaystyle \pi\frac{R^2}{H^2}h^2 dh$. We can find the volume between two different height, $\displaystyle h_1$ and $\displaystyle h_2$ by integrating that:
    $\displaystyle \frac{R^2}{H^2}\pi\int_{h_1}^{h_2} h^2dh= \frac{\piR^2}{3H^2}(h_2^3- h_1^3)$.

    So the distance from $\displaystyle h_1= 0$, at the vertex, to the height, $\displaystyle h_2$ where the volume is 1000 ml is $\displaystyle \frac{\pi R^2}{3H^2}h_2^3= 1000$. Once you have solved for that h, set it equal to $\displaystyle h_1$ and solve for the next $\displaystyle h_2$.
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  3. #3
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    Unfortunately I had worked out the previous question that makes you find the function for the volume of the cone. Instead of $\displaystyle \frac {32a^3}{27}$ it should actually be $\displaystyle \frac {32\pi a^3}{81}$
    Last edited by FLOTUS; Aug 14th 2010 at 10:09 PM.
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