Help solving volume equation

Question

I'm working on some ecological data that uses the following equation to calculate volume (V) of timber logs in cubic metres per hectare based on surveys along a transect.

$$ V=\frac{\pi^2\Sigma d^2}{8L}$$

where L is length of transect, and d is diameter of the log.

My question is do square all the individual diameters, then add them, then add $\pi^2$ or do I sum all diameters, square them, then add $\pi^2$?

Bonus points for explanations! Thanks in advance.

Answer 1

Remember the order of operations:

1. parentheses

2. exponents

3. multiplication and division from left to right, with multiplication and division (which is multiplication by the multiplicative inverse of the divisor) taking equal precedence

4. addition and subtraction from left to right, with addition and subtraction (which is addition of the additive inverse of the subtrahend) also taking equal precedence

The summation symbol acts like a set of parentheses, so we do what is inside first. Recall that

$$\sum_{k = 1}^n a_k = a_1 + a_2 + \cdots + a_n$$

In your example, $a_k = d_k^2$, so we obtain

$$\sum_{k = 1}^n d_k^2 = d_1^2 + d_2^2 + \cdots + d_n^2$$

Since exponentiation takes precedence over addition, we first square each $d_k$, then add. If we wanted to represent the square of the sum, we would instead write

$$(d_1 + d_2 + \cdots + d_n)^2 = \left(\sum_{k = 1}^n d_k\right)^2$$

Once we have performed the addition, we have a product of three factors:

\begin{align*} \frac{\pi^2\sum_{k = 1}^n d_k^2}{8L} & = \pi^2 \cdot \sum_{k = 1}^n d_k^2 \cdot \frac{1}{8L} && \text{since division is multiplication by the reciprocal}\\ & = \pi^2 \cdot \frac{1}{8L} \cdot \sum_{k = 1}^n d_k^2 && \text{since multiplication is commutative}\\ & = \frac{\pi^2}{8L}\sum_{k = 1}^n d_k^2 && \text{multiply from left to right} \end{align*}

In answer to your question, we first square each $d_k$, add the squares, then multiply the sum by $\pi^2$, and divide the result by $8L$. Note that multiplying by $\pi^2$, then dividing by $8L$ is equivalent to multiplying by $\dfrac{\pi^2}{8L}$.

Answer 2

I can't speak as to the validity of the equation, but what it means in this form is to square each of the $d$s, add them, and then multiply by $\dfrac{\pi^2}{8L}$. So for example if you had $d_1=1$, $d_2=2$, and $d_3=3$ for $3$ different logs, you would get

$$\frac{\pi^2}{8L}\times\left(1^2+2^2+3^2\right)$$

Answer 3

From what I can guess the question is how to compute $$\pi^2 \sum_{i=1}^N d_i^2 = \pi^2 (d_1^2 + d_2^2 + d_3^2 + \ldots + d_N^2)$$ This means we first sum the squares of all diameters and then multiply the result (not add) by $\pi^2$.