Answer
(a) $V(d) = \frac{d^3 \pi} 6$
(b) \begin{matrix}
d & V(d) \\
0 & 0 \\
1 & 0.524 \\
2 & 4.189 \\
3 & 14.137 \\
4 & 33.51
\end{matrix}
(c) Image below:
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Work Step by Step
(a) $V(d) = ?$
- Take the cube of the diameter
$V(d) = d^3 ...$
- Multiply by $\pi$:
$V(d) = d^3 \pi...$
- Divide by 6:
$V(d) = \frac{d^3 \pi} 6$
(b) Make a table of values of $V(d)$. You can choose any values for $d$. I will use 0, 1, 2, 3 and 4.
- Calculate $V(0)$, $V(1)$, $V(2)$, $V(3)$ and $V(4)$:
$$V(0) = \frac{(0)^3 \pi} 6 = 0$$ $$V(1) = \frac{(1)^3 \pi} 6 = 0.524$$ $$V(2) = \frac{(2)^3 \pi} 6 = 4.189$$ $$V(3) = \frac{(3)^3 \pi} 6 = 14.137$$ $$V(4) = \frac{(4)^3 \pi} 6 = 33.51$$
- Make the table: \begin{matrix}
d & V(d) \\
0 & 0 \\
1 & 0.524 \\
2 & 4.189 \\
3 & 14.137 \\
4 & 33.51
\end{matrix}
(c) Use the table we made to plot the points with the respective coordinates. Then sketch a curve that passes through all these points.