Area, Surface Area and Volume: Question 8

Worked solution

Part (a): Radius and volume of the hemisphere

The diameter of the cylinder is $12\text{ cm}$, and the hemisphere sits on top with the same diameter. The radius is half the diameter:

$r = \frac{12}{2} = 6\text{ cm}.$

A hemisphere is half of a sphere, so its volume is half of $\frac{4}{3}\pi r^3$:

$V_{\text{hemi}} = \frac{1}{2}\times\frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3.$

Substitute $r = 6$:

$V_{\text{hemi}} = \frac{2}{3}\pi (6)^3 = \frac{2}{3}\pi \times 216 = 144\pi\ \text{cm}^3.$

Answer: $r = 6\text{ cm}$ and $V_{\text{hemi}} = 144\pi\ \text{cm}^3$.

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Part (b): Total volume of the trophy

The trophy is a cylinder with a hemisphere on top, so add the two volumes.

Volume of the cylinder uses $V = \pi r^2 h$ with $r = 6\text{ cm}$ and $h = 21\text{ cm}$:

$V_{\text{cyl}} = \pi (6)^2 (21) = \pi \times 36 \times 21 = 756\pi\ \text{cm}^3.$

Add the hemisphere from part (a):

$V_{\text{total}} = V_{\text{cyl}} + V_{\text{hemi}} = 756\pi + 144\pi = 900\pi\ \text{cm}^3.$

Evaluate to 3 significant figures. Using $\pi = 3.142$ (or the calculator value):

$V_{\text{total}} = 900 \times 3.142 = 2827.8\ldots \approx 2830\ \text{cm}^3\ (3\text{ s.f.})$

(The more precise calculator value is $900\pi = 2827.43\ldots$, which also rounds to $2830\text{ cm}^3$.)

$\boxed{V_{\text{total}} = 900\pi \approx 2830\ \text{cm}^3}$

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Quick check

Working in terms of $\pi$ first ($756\pi + 144\pi = 900\pi$) keeps the arithmetic exact and avoids rounding errors; only convert to a decimal at the very end. Note also that the hemisphere ($144\pi$) is smaller than the tall cylinder ($756\pi$), which makes sense as the cylinder is the bulk of the solid.