1. ## 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

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

Any help would be greatly appreciated

2. 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 $\frac{\pi}{3}a^2h$. You say that the volume is given by $\frac{32a^3}{27}$ which leads me to the conclusion that the height is $\frac{a}{9\pi}$. With a= 9 cm, that gives $h= \frac{9}{9\pi}= \frac{1}{\pi}$ 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 $\frac{r}{h}= \frac{R}{H}$ or $r= \frac{R}{H}r$. A circle with that radius has area $\pi r^2= \pi\frac{R^2}{H^2}h^2$ and, if we think of it as a thin disk of height dh, volume $\pi\frac{R^2}{H^2}h^2 dh$. We can find the volume between two different height, $h_1$ and $h_2$ by integrating that:
$\frac{R^2}{H^2}\pi\int_{h_1}^{h_2} h^2dh= \frac{\piR^2}{3H^2}(h_2^3- h_1^3)$.

So the distance from $h_1= 0$, at the vertex, to the height, $h_2$ where the volume is 1000 ml is $\frac{\pi R^2}{3H^2}h_2^3= 1000$. Once you have solved for that h, set it equal to $h_1$ and solve for the next $h_2$.

3. Unfortunately I had worked out the previous question that makes you find the function for the volume of the cone. Instead of $\frac {32a^3}{27}$ it should actually be $\frac {32\pi a^3}{81}$