"Remember the year that we spent as that pool of slime?"
"Spread out on that rock on the mountainside?"
"Yeah, that was nice."
"Just lying there under the sun, feeling the texture of the rock."
"We got to know it really well."
"And how the sun and the rock and the temperature gradient made our top surface different from our bottom surface."
"And the layer in between. And the microbes!"
"Yeah, the microbes always swimming around inside us, eating and reproducing and all."
"Those were the days."
"You said it."
So when you have poison ivy, do you get this strange shuddery frisson, somehow ecstatic or mildly orgasmic, when you hold the irritated area under very hot water? Does this count as scratching (which, under Poison Ivy Rules, makes it get worse)? Or is it Safe?
The other night (I was perhaps a bit hyper) I was talking about solid geometry and calculus and stuff with the little daughter, and not being able to remember the formula for the volume of a right circular cone, I proposed to derive it. And in fact to derive it in two different ways.
(The little daughter thought this was sort of odd and amusing, and didn't stay for the whole thing, although we did have some discussion of the various methods over dinner later on.)
As a big grownup professional research person and all with a thorough technical and liberal education this should really be beneath my notice, but it was fun, so let's do it.
The most obvious method is to put the cone nose-down on the origin, with its centerline on the sticking-up ("z") axis. (Ya know, this would be much easier to do if I were willing to make some pictures.) Call the cone's height H, and the radius of the base R.
Now we'll add up the areas of the infinite number of circles described by the intersection of the cone with a flat plane at some height h over the origin (the plane z=h for some h between 0 and H).
What's the radius of each of these circles? Well, the sides of the cone are nice straight lines, and the radius is zero at h=zero and R at h=H, so at h it's clearly (R/H) times h. And therefore the area of the circle (which is pi times the radius squared) must be pi times (R-squared over H-squared) times h-squared, or in funny symbols π(R^2/H^2)h^2.
Integrating that dh, we get (π(R^2/H^2)h^3)/3. That's zero at h=0, so the answer we want is what it is at h=H, which is πR^2H/3.
Ha, wild! So it's just one third of the area of the cylindrical box that it comes in. Who'd'a thought?
The second way we considered figuring it was by summing up the triangles formed by slicing the cone with vertical planes (x=r, say). But the formula for the bases of those triangles looked like it was going to have sines and/or cosines and/or things in it, and we turn out to be very lazy.
In a burst of inspiration I decided to sum up the surfaces of the cylinders formed by intersecting the cone with cylinders that are concentric with it and have various radii. (For simplicity we'll also turn the cone over, so it's sitting on the origin dunce-cap-wise.) The innermost such slice has height H and radius 0 (and therefore surface area zero), and the outermost one has height 0 and radius R (and therefore also surface area 0).
For a cylinder in between, with radius r (0<r<R), the circumference is 2πr, and the height is (R-r)(H/R), again because the cone has nice straight sides. So the surface area is just 2π(H/R) times r times (R-r), or 2π(H/R)r(R-r), or more to the point 2πHr-2πHr^2/R.
Integrating that mess dr, we get πHr^2 - (2πHr^3)/(3R). That's conveniently zero when r is zero again, so we just need to evaluate it at r=R, which gives πHR^2 - 2πHR^2/3, which is πHR^2/3 or most gratifyingly πR^2H/3 again.
Yay, us!
We can still do utterly elementary calculus even after all these years in digital-land!
*8)
