Exercises — Surface area and volume of all above 3D shapes

Volume is a stack of square slices, each of area $a^2$, piled to height $a$: $V=a^3=4^3=64{\text{cm}}^3$ Surface area is 6 identical square faces: $SA=6a^2=6\times 16=96{\text{cm}}^2$ The units differ on purpose: volume is ${\text{cm}}^3$ (three lengths multiplied), area is ${\text{cm}}^2$ (two lengths). See Dimensional Analysis.

The formula uses radius, not diameter. Since $d=2r$, we halve it: $r=10/2=5\text{cm}$. $V=\pi r^{2}h=\pi \times 5^2\times 6=150\pi {\text{cm}}^3\approx 471.2{\text{cm}}^3$

Step 1 — find the slant height $l$. Look at the red right triangle in the figure: the vertical leg is the height $h=8$, the horizontal leg is the radius $r=6$, and the sloping side (hypotenuse) is $l$. By the Pythagorean Theorem the two legs squared add to the hypotenuse squared: $l=\sqrt{r^2+h^2}=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10\text{cm}$ Why this tool? Pythagoras is the only rule that turns two perpendicular lengths into the diagonal between them — exactly what a slant is. Step 2 — total surface area = base circle + curved sector: $TSA=\pi r(r+l)=\pi \times 6\times (6+10)=96\pi {\text{cm}}^2\approx 301.6{\text{cm}}^2$

Step 1 — undo the surface formula for $r$. We know $SA=4\pi r^2$. Set it equal and solve: $4\pi r^2=100\pi \Rightarrow r^2=\frac{100\pi }{4\pi }=25\Rightarrow r=5\text{cm}$ Why divide by $4\pi$? To isolate $r^2$ we peel off everything multiplying it. We take the positive root only, because a radius cannot be negative. Step 2 — plug into the volume formula: $V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi \times 125=\frac{500}{3}\pi {\text{cm}}^3\approx 523.6{\text{cm}}^3$

The solid is two shapes glued at a shared circular seam. Volumes simply add — no piece is lost, no piece is double-counted, because the flat top of the cylinder is the flat base of the hemisphere and it lies inside the solid. Cylinder part: $V_{\text{cyl}}=\pi r^{2}h=\pi \times 9\times 10=90\pi {\text{m}}^3$ Hemisphere part (half a sphere): $V_{\text{hemi}}=\frac{2}{3}\pi r^3=\frac{2}{3}\pi \times 27=18\pi {\text{m}}^3$ Total: $V=90\pi +18\pi =108\pi {\text{m}}^3\approx 339.3{\text{m}}^3$

Look again at the figure. Three surfaces are exposed:

• Cylinder curved wall: $2\pi rh=2\pi \times 3\times 10=60\pi {\text{m}}^2$
• Cylinder floor (bottom circle): $\pi r^2=9\pi {\text{m}}^2$
• Hemisphere dome (curved half only, not its flat disc — that disc is buried against the cylinder top): $2\pi r^2=18\pi {\text{m}}^2$ We deliberately skip the flat top of the cylinder and the flat base of the hemisphere: they touch each other on the inside, so no paint reaches them. $SA=60\pi +9\pi +18\pi =87\pi {\text{m}}^2\approx 273.3{\text{m}}^2$

The key idea: melting changes shape but conserves volume. So $V_{\text{cone}}=V_{\text{sphere}}$. Cone volume: $V_{\text{cone}}=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi \times 36\times 24=288\pi {\text{cm}}^3$ Set equal to sphere volume and solve for the new radius $R$: $\frac{4}{3}\pi R^3=288\pi \Rightarrow R^3=\frac{288\times 3}{4}=216\Rightarrow R=\sqrt[3]{216}=6\text{cm}$ Why cube root? The formula gave $R^3$; to free the single length $R$ we undo the cube with a cube root. $6^3=216$, so $R=6\text{cm}$. (See Density and Mass — same mass, same volume, only the shape moved.)

Volume depends on $a^3$. Replace $a$ by $2a$: $V_{\text{new}}=(2a{)}^3=8a^3\Rightarrow \text{volume }\times 8$ Surface area depends on $a^2$. Replace $a$ by $2a$: $S{A}_{\text{new}}=6(2a{)}^2=24a^2\Rightarrow \text{surface area }\times 4$ Why the mismatch? Volume grows with the cube of any length scale, area with the square. Doubling length gives $2^3=8$ and $2^2=4$. This is the heart of Similar Solids: scale by $k$, then area $\times k^2$, volume $\times k^3$.

Subtract the hole from the block. The drilled cylinder shares the block's height $h=4$. Block volume: $V_{\text{block}}=lwh=10\times 6\times 4=240{\text{cm}}^3$ Hole volume (a cylinder, $r=1$, height $4$): $V_{\text{hole}}=\pi r^{2}h=\pi \times 1\times 4=4\pi {\text{cm}}^3$ Remaining: $V=240-4\pi \approx 240-12.57=227.4{\text{cm}}^3$ Notice we cannot combine the $240$ and the $4\pi$ into one clean number — one is a whole integer, the other carries $\pi$. Leave it as $240-4\pi$.

Step 1 — what happens geometrically. With $l$ fixed at $5$ and $h\to 0$, solve for $r$: $r=\sqrt{l^2-h^2}=\sqrt{25-h^2}\stackrel{h\to 0}{\to }\sqrt{25}=5\text{cm}$ The apex sinks down onto the base plane. The cone flattens into a flat disc of radius $5\text{cm}$ — a degenerate cone with no height at all. Step 2 — curved surface area in the limit. The curved (lateral) area is $\pi rl$. As $h\to 0$, $r\to 5$ and $l=5$: $CSA\to \pi \times 5\times 5=25\pi {\text{cm}}^2\approx 78.5{\text{cm}}^2$ Sanity check: a fully flattened cone's "curved" surface presses onto a disc of radius $5$, whose area is $\pi r^2=25\pi$. The two agree — the geometry is consistent right up to the boundary case. (See Integration and Calculus for why limits like this are trustworthy.)

Sphere: we're told the volume equals $\frac{4}{3}\pi r^3$ with $r=3$, so $r=3\text{cm}$ and $S{A}_{\text{sphere}}=4\pi r^2=4\pi \times 9=36\pi \approx 113.1{\text{cm}}^2$ Cube: find the edge $a$ from equal volume. Volume $=36\pi \approx 113.097{\text{cm}}^3$, so $a=\sqrt[3]{36\pi }\approx \sqrt[3]{113.097}\approx 4.837\text{cm},S{A}_{\text{cube}}=6a^2\approx 6\times 23.40\approx 140.4{\text{cm}}^2$ Conclusion: the sphere needs $\approx 113.1{\text{cm}}^2$ of skin, the cube $\approx 140.4{\text{cm}}^2$ — the sphere wins by roughly $27{\text{cm}}^2$. For a fixed volume, the sphere has the smallest possible surface area, which is why bubbles, droplets, and planets are round. See Real-World Applications.

The volume of a sphere is ==$\frac{4}{3}\pi r^3$ and its surface area is $4\pi r^2$==. Scaling every length of a solid by $k$ multiplies its surface area by $k^2$ ::: and its volume by $k^3$. At a glued seam between two solids, how many of the two touching faces get painted? ::: Zero — both are hidden inside. To find a cone's slant height from $r$ and $h$ you use ::: the Pythagorean theorem, $l=\sqrt{r^2+h^2}$.