Take a cup of water. How do you calculate the volume? Good question. Glad you asked. Fortunately for us, we have this thing called calculus.
All you have to do is find a mathematical model that describes the shape of the cup, integrate, and voila, you have the volume.
Now, a standard pixie cup can be represented by a truncated circular cone, which can be described via:
The cross-section of the cup, going through the xy plane, is a circle, described by:
With a = z for any particular cross section. The area of a circle is... as we all know:
Since the radius is z for our example, the area of a cross section becomes:
Great, so we've got the area for each cross section of our pixie cup. Now, to calculate the volume, we merely integrate the area of each cross-section over the height of the cup. Simple.
Now, there's the fact that c <= z <= d, where c is the radius of the cup at the bottom and d is the radius of the cup at the top. Hence, the resulting integral becomes:
The beautiful thing about calculus is that there are various ways to visualize things. For example, instead of integrating from c to d, we could calculate two integrals from 0 to c and 0 to d, respectively, then subtract the former from the latter. This has the effect of finding the volume of a large circular cone and subtracting the volume of a smaller cone from it. This looks like:
Voila.
Up until now we've been using arbitrary constants. To find the volume for this particular cup, all we need to do is make a couple of measurements.
Ah, we find that the bottom radius is 1.5 inches and the top radius is 2 inches.
Plugging these numbers into our formula, we see that the cup contains 24.74 cubic inches of water. Now, this isn't terribly useful, so let's convert to cups.
The ratio of cubic inches to cups is 1 cubic inch to 0.06926 cups. So... multiplying by 0.06926, we have 1.71349 cups.
...or you could just use a measuring cup. That works too. Sometimes using math to solve certain problems is analogous to hunting moose with bazookas.
Until next time,
- Daniel
Or you could find the mass of the water and the density of water (4.184 J)- Divide the mass by the density and you get the volume... or you could just use the measuring cup :)
ReplyDeleteI like. The problem comes when your water doesn't come in cones... when it comes in oblongy sorts of things which are sorta round but don't go all the way around. :P Think 1/6ths of paraboloids.
ReplyDeleteActually, it's easier to hunt moose (meese? mooses? mice?) with a bazooka. There's a lot more margin for error.
ReplyDeleteLoving the calculus. Loving the fact that you posted about calculus on your blog. Loving the opening and closing.
ReplyDeleteThere's just one problem. The model you chose. You have x^2 + y^2 - z^2 = 0. OK, yeah, a circular cross-section with... quadratically... increasing radius...
Your safest model for a general truncated cone is x^2 + y^2 - k*z = 0. Linearly increasing radius, unknown constant defining just how quickly that radius increases.
Then the volume of each slice is pi*r^2*dz = pi*(kz)^2*dz = pi*k^2*z^2 dz. (Around now I find it is time to regret that I don't know a way of putting these equations in a pretty form in a comment.) The indefinite integral is then 1/3*pi*k^2*z^3 + C.
To find the appropriate bounds of integration, you need to figure out what z is; since k is now an additional variable, you'll need not only the two radii but also the height. Then you can find k, z1, and z2 by setting k*z1 = r1, k*z2 = r2, and z2 - z1 = h, and solving the three equations in three variables. Then your final answer is 1/3*pi*k^2*(z2^3 - z1^3).
Oh, let's put that in terms of r1, r2, and h... z1 = r1/k, z2 = r2/k, so (r2-r1)/k = h, so r2 - r1 = kh, so k = (r2-r1)/h. Then z1 = r1*h/(r2-r1) and z2 = r2*h/(r2-r1). So your final answer for the volume of a Dixie cup from the radii and the height is... 1/3*pi*(r2-r1)^2/h^2*(r2^3-r1^3)*h^3/(r2-r1)^3. Cancelling... the answer is 1/3*pi*h*(r2^3-r1^3)/(r2-r1). If I did that all right.
Yeah, you might want to pick a non-calculus solution... :) ...but they're much less precise, and they won't let you do fun things like progressing to optimizing how much volume you get for a given amount of cup surface area! :)
This is Melanie, by the way, commenting as, well, my one blog, which is a group blog, thus not under my name. I see no option for commenting under just a name, thus...
Agh. Kindly allow me to eat my words. Of course it's r^2, not r, which x^2 + y^2 must equal; z^2 is totally correct; duh. Remind me to think things through better before trying to correct intelligent people, heh. It *would* be a good idea to add in a k, however, rather than assuming that k=1 (perfect 45 degree angle sides, which I don't *think* is normal for a styrofoam cup). Which will get the height involved. The answer should reduce to 1/3*pi*h*r^2 for the full cone.
ReplyDelete-Melanie