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Well and embankment volume: model the ring before calculating

7 min read

Question

A cylindrical well has radius 4 m and depth 15 m. The earth dug out is spread evenly to form a 2 m wide embankment around the well. If all the earth is used, find the height of the embankment.

Well & Embankment: Why 'earth dug' isn't always 'embankment volume'Watch on YouTube

The embankment is 12 m high. The arithmetic is short once the shape is modelled correctly: earth comes from a cylindrical well, but it is spread into a hollow cylindrical ring around that well.

outer R = 6 mhole filledSolid cylinderfull disc baseh = 6.67 mThis counts soil inside the well.
π(62)h=240π\pi\left(6^2\right)h=240\piSolving gives h = 6.67 m, which is too small.
A solid outer cylinder quietly puts earth back inside the well. Tap the button to repair the model.

Step 1: find the volume of earth dug out

The well is a cylinder with radius 4 m and depth 15 m:

Vwell=πr2dV_{\text{well}}=\pi r^2d
Vwell=π(4)2(15)V_{\text{well}}=\pi(4)^2(15)
Vwell=π(16)(15)V_{\text{well}}=\pi(16)(15)
Vwell=240π m3V_{\text{well}}=240\pi\text{ m}^3

Step 2: model the embankment as a ring

First find the outer radius. Then, if the unknown height is hh, subtract the inner cylinder from the outer one:

R=r+widthR=r+\text{width}
R=4+2R=4+2
R=6 mR=6\text{ m}
Vembankment=π(R2−r2)hV_{\text{embankment}}=\pi(R^2-r^2)h
Vembankment=π(62−42)hV_{\text{embankment}}=\pi(6^2-4^2)h
Vembankment=π(36−16)hV_{\text{embankment}}=\pi(36-16)h
Vembankment=20πhV_{\text{embankment}}=20\pi h

Step 3: use conservation of volume

In the school-model version of the problem, all the earth removed from the well becomes the embankment. Equate the two volumes:

Vembankment=VwellV_{\text{embankment}}=V_{\text{well}}
20πh=240π20\pi h=240\pi
20h=24020h=240
h=24020h=\frac{240}{20}
h=12 m\boxed{h=12\text{ m}}
area ratio=20π16π\text{area ratio}=\frac{20\pi}{16\pi}
area ratio=54\text{area ratio}=\frac54
h=15÷54h=15\div\frac54
h=15×45h=15\times\frac45
h=12 mh=12\text{ m}

Why the tempting solid-cylinder method fails

Using π(6)2h\pi(6)^2h treats the embankment as if earth also occupied the well opening. That calculation gives about 6.67 m, but it solves a different physical situation.

π(6)2h=240π\pi(6)^2h=240\pi
36πh=240π36\pi h=240\pi
36h=24036h=240
h=24036h=\frac{240}{36}
h=203 mh=\frac{20}{3}\text{ m}
h≈6.67 m(wrong model)h\approx6.67\text{ m}\quad\text{(wrong model)}

A compact formula for this entire problem type

If a well of radius rr and depth dd creates an embankment of widthww and height hh, then R=r+wR=r+w. Conservation of volume gives:

πr2d=π((r+w)2−r2)h\pi r^2d=\pi\left((r+w)^2-r^2\right)h
r2d=((r+w)2−r2)hr^2d=\left((r+w)^2-r^2\right)h
r2d=(r2+2rw+w2−r2)hr^2d=\left(r^2+2rw+w^2-r^2\right)h
r2d=(2rw+w2)hr^2d=\left(2rw+w^2\right)h
r2d=w(2r+w)hr^2d=w(2r+w)h
h=r2dw(2r+w)\boxed{h=\frac{r^2d}{w(2r+w)}}

Exam checklist

  • Sketch the top view and mark both radii.
  • Write R=r+wR=r+w before substituting numbers.
  • Use π(R2−r2)h\pi(R^2-r^2)h for the ring.
  • Equate volumes only when the question says all the earth is used.
  • Keep every length in the same unit before calculating.

Real soil can expand when loosened or compact when packed. Unless a question supplies an expansion or compaction factor, the intended mathematical model assumes equal volumes.

Build the ring, change its dimensions, and practise the surface-areas-and-volumes lesson in Young Math Brains.

Practise composite volumes interactively

Your turn

Why must the well's circular opening be subtracted when you calculate the embankment volume?

Tell us in the comments on YouTube or Instagram, or email connect@youngmathbrains.com. We read every one.

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