Calculus question for top contenders?

Question

This is a question given to me by an ex WWII engineer a few years back as a challenge to me. It took me quite some time to figure out. Thought it might be found an interesting question and challenge.

If you have a cylinder of radius r and with a drill of the same radius you drill it so that it JUST breaks in half. How much volume of material have you removed?

you are drilling perpendicular to the cylinder so that it breaks in two.

Answer 1

Are you drilling perpendicular to the axis of the cylinder or at an arbitrary angle?

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After napping for a bit I came back to see that impressive manipulation below (!) -- but it can be done a lot more simply.

By looking at the symmetry of the intersection of the two cylinders, imagine splitting it into eight chunks of the same shape -- an r by r square on the back face aligned with the axis of the cylinder with curved surfaces formed by two circular arcs sweeping to the point where they meet at the edge of the cylinder.

This thing can be cut up into little *square* slices of width dx. The biggest one is r by r by dx and they progressively shrink down to nothing when x = r. In the middle they have sides given by sqrt(r^2 - x^2).

So the integration turns out to be really simple. It's just

int((r^2 - x^2)dx, 0 .. r)

= (r^2 x - x^3/3) | 0 .. r

= r^3 - r^3/3

= 2/3r^3.

But this chunk is only 1/8th of the whole thing, so the volume removed is

16/3 r^3.

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Thanks, Pascal -- now I'd like to see a solution to the harder problem of drilling at an arbitrary angle. I've been daydreaming a bit about it but haven't made a lot a progress.

Answer 2

We will let the cylinder be centered at the origin and the axis aligned in the y direction. We will drill into the origin from the positive z-direction.

Note first that at a point (x, y) in xy-plane, the cylinder runs from a z-height of -√(r² - x²) to √(r² - x²), so the height of material removed is 2√(r² - x²). The drill removes all the material directly above or below the disk D = {(x, y): x² + y² ≤ r²}. So the volume of material removed is:

[D]∬2√(r² - x²) dA

Solving this integral, let us find the limits of integration. They are -r≤y≤r, -√(r² - y²)≤x≤√(r² - y²). Thus we have:

[-r, r]∫[-√(r² - y²), √(r² - y²)]∫2√(r² - x²) dx dy

To make some computations regarding sign easier, we will use the fact that [-√(r² - y²), √(r² - y²)]∫2√(r² - x²) dx = 2 [0, √(r² - y²)]∫2√(r² - x²) dx, since 2√(r² - x²) is an even function. Similarly, 2 [0, √(r² - y²)]∫2√(r² - x²) dx is an even function, so by symmetry, [-r, r]∫2 [0, √(r² - y²)]∫2√(r² - x²) dx dy = 2 [0, r]∫2 [0, √(r² - y²)]∫2√(r² - x²) dx dy. Thus we have:

2 [0, r]∫2 [0, √(r² - y²)]∫2√(r² - x²) dx dy

Collecting the constant factors

8 [0, r]∫[0, √(r² - y²)]∫√(r² - x²) dx dy

Extracting a factor of r:

8r [0, r]∫[0, √(r² - y²)]∫√(1 - (x/r)²) dx dy

So now make the substitution θ = arcsin (x/r), so sin θ = x/r, x = r sin θ, and dx = r cos θ dθ. Now to find the new limits of integration, note that when x=0, θ=0, and when x=√(r² - y²), we have sin θ = x/r = √(1 - (y/r)²), so sin² θ = 1 - (y/r)², thus (y/r)² = 1 - sin² θ = cos² θ, thus cos θ = ±y/r. But since the range of arcsin is [-π/2, π/2], cos θ is positive, and since y∈[0, r] is positive, we in fact have cos θ = y/r. So then θ = ±arccos (y/r). However, since x is positive, arcsin (x/r) is positive, and as arccos (y/r) is positive, we in fact have θ = arccos (y/r). So we have:

8r [0, r]∫[0, arccos (y/r)]∫√(1 - sin² θ) r cos θ dθ dy

Simplifying this a bit:

8r² [0, r]∫[0, arccos (y/r)]∫√(cos² θ) cos θ dθ dy
8r² [0, r]∫[0, arccos (y/r)]∫cos² θ dθ dy (since cos θ is positive)

Now, using the formula cos² θ = (cos (2θ) + 1)/2:

4r² [0, r]∫[0, arccos (y/r)]∫cos (2θ) + 1 dθ dy

Integrating:

4r² [0, r]∫1/2 sin (2θ) + θ |[0, arccos (y/r)] dy
4r² [0, r]∫1/2 sin (2 arccos (y/r)) + arccos (y/r) dy

Simplifying a bit:

4r² [0, r]∫sin (arccos (y/r)) cos (arccos (y/r)) + arccos (y/r) dy
4r² [0, r]∫sin (arccos (y/r)) y/r + arccos (y/r) dy
4r² [0, r]∫√(1 - cos² (arccos (y/r))) y/r + arccos (y/r) dy (since arccos (y/r) is positive, so is sin (arccos (y/r)), thus the sign is correct)
4r² [0, r]∫√(1 - (y/r)²) y/r + arccos (y/r) dy

Breaking this into two integrals:

4r² [0, r]∫√(1 - (y/r)²) y/r dy + 4r² [0, r]∫arccos (y/r) dy

Simplifying the first:

2 [0, r]∫√(r² - y²) 2y dy + 4r² [0, r]∫arccos (y/r) dy

Making the substitution u=r² - y², we have that du = -2y dy, and that when y=0, u=² and when y=r, u=0. So we have:

-2 [r², 0]∫√u du + 4r² [0, r]∫arccos (y/r) dy

Integrating:

-4/3 u^(3/2)|[r², 0] + 4r² [0, r]∫arccos (y/r) dy
4/3 r³ + 4r² [0, r]∫arccos (y/r) dy

Now, turning our attention to the second integral, we proceed by parts. Let u=arccos (y/r) and dv = dy, du = -1/√(1 - (y/r)²) * 1/r dy and v=y. So we have:

4/3 r³ + 4r² (y arccos (y/r)|[0, r] + [0, r]∫y/√(1 - (y/r)²) 1/r dy)

Simplifying:

4/3 r³ + 4r² (r arccos (1) + [0, r]∫y/√(r² - y) dy)
4/3 r³ + 4r² [0, r]∫y/√(r² - y²) dy (since arccos (1) = 0)

Making the substitution u = r² - y², we have (again) that du = -2y dy, y=0 ⇒ u=r², and y=r ⇒ u=0, so we have:

4/3 r³ - 2r² [r², 0]∫1/√u du

Integrating:

4/3 r³ - 4r²√u|[r², 0]
4/3 r³ + 4r³
16/3 r³

And we are done.

Edit: corrected arithmetic error near the end. Searching for other possible errors -- give me a minute.

Edit 2: Found a second error, and corrected it.

Edit 3: Okay, this derivation is now correct. Regarding the answers by mr_zone_v and kendizzle -- the shape of the removed material is neither a cylinder nor a sphere. Rather, it is the intersection of two cylinders, which is known as a Steinmetz solid. Thus any volume calculations based on the assumption that it is a sphere or cylinder will be quite wrong.

Edit 4: Wow steve -- that's brilliant! Wish I'd thought of that -- it would have saved me a lot of time.

Answer 3

4/3 pi r^^3

The cylinder is 3-dimensional, and let's say its x and y axes are in the plane of its circular cross section, while its z axis goes along the height of the cylinder. Let's then say we drill straight down the x axis, which is perpendicular to the z axis. The drill removes material by boring a cylindrical path down x, but we can envision this boring cylinder as a succession of circular y-z planes stacked along the x axis. That means we have subtended circles in the x-y plane of radius 'r' (ths cylinder's cross-section), and the y-z plane (the drill's bore, also 'r'). So from the unique point in the cylinder where the z axis and x (drilling) axis intersect, we remove material in 3 dimensions to a distance r -->> the shape of the material removed is a sphere, so the volume of the material removed is 4/3 pi r^^3

Answer 4

doesn't it depend on the pitch of the drill bit's tip.

if the tip was extremely pointy than there would be a hollowed out part at the top of the remaining half, so you would have to subtract that cavity's volume from the volume of the cylinder.

I don't think there is a definite answer with the info given.

After reading other responses I assume you mean drilling perpendicular to the cylinders axis.

in that case the answer is easy:

becuase when you drill perpendicular to the axis of the cylinder you create a half cylinder cavity at one of the ends.

the dimensions of that cavity are:
radius = r/2
height = 2r
volme = (pi/2) * (r/2)^2 * (2r) = (pi/16) * r^3

the dimensions of the cylinder removed are:
radius = r
height = h/2
volume = (pi/2) * r^2 * h

add the two and you get:
(pi/16)r^3 + (pi/2)r^2h

Answer 5

Could you elaborate on the question? I assume you mean the cylinder breaks in half, but what does that mean? Break in half vertically or horizontally? Does it break because the drill reaches the bottom of the cylinder? I'm just not too sure on all the information that's given/needed.

Answer 6

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Answer 7

Half of the volume

Answer 8

you get the same volume as before
V = (pi) * r^2 * h

(pi) is still the same
radius is still the same
and height is still the same

therefore nothing changed.....
???