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I am trying to calculate the volume of some burial mounds, but I am embarrassingly poor at math.

The shape of an ideal burial mound is most similar to a hemisphere. I have read other people's work and they use this formula, but I can't really get it to work:

$$V = \pi H \left(\left(\frac{R^2}{2} + \frac{H^2}{6}\right) + \frac{H^2}{6}\right)$$

I only have the height and the diameter of the burial mounds, so the R is radius and H is height. For example, one burial mound is $30m$ in diameter and $6m$ high. The volume should be $2218m^3$ according to another author. How do I calculate this?

conker
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  • I wasn't entirely sure how to interpret "R2" and "H2" in the original, so feel free to fix the equation if they were intended as unsimplified fractions, rather than subscripts. –  Jun 19 '15 at 07:24
  • To make units work out, I'd interpret R2 as $R^2$ and H2 as $H^2$. – JimmyK4542 Jun 19 '15 at 07:25
  • Silly me! That would have been a smart thing to check first. I adjusted the parentheses to get a value closer to the expected result. –  Jun 19 '15 at 07:27
  • See https://en.wikipedia.org/wiki/Spherical_segment and https://en.wikipedia.org/wiki/Spherical_cap – CiaPan Jun 19 '15 at 07:33

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The formula "V=π H(R2/2 + H2/6)/2 + H2/6)", is hard to read, and has a parenthesis imbalance, i.e. there are two left parenthesis and only one right parenthesis.

You mentioned that the ideal shape is similar to a hemisphere. Since the height isn't exactly half of the diameter, the mound isn't exactly a half of a sphere, but rather a spherical cap.

The volume of a spherical cap with base radius $R$ and height $H$ is $V = \dfrac{\pi H}{6}\left(3R^2+H^2\right)$.

Here, the base has radius $a = \dfrac{30 \text{m}}{2} = 15\text{m}$ and the height is $h = 6\text{m}$.

Hence, the volume is $V = \dfrac{\pi H}{6}\left(3R^2+H^2\right) = \dfrac{\pi (6 \text{m})}{6}\left(3(15\text{m})^2+(6\text{m})^2\right) \approx 2234 \text{m}^3$, which is close to what the author got.

JimmyK4542
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