Visual walkthrough — Surface area and volume of all above 3D shapes

Everything here rests on things you can already picture: a stack of coins, a fan of pizza slices, and one right triangle. We build the rest.

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Step 1 — What even is a cone? Name the three lengths

WHAT. A cone is what you get if you take a circle of radius $r$, lift a single point (the apex) straight up to a height $h$ above the circle's centre, and draw straight lines from that apex down to every point on the circle's rim.

WHY start here. You cannot measure a shape you cannot name. Three lengths control everything:

• $r$ = radius of the circular base (how wide the bottom is),
• $h$ = height (how high the apex sits straight up from the centre),
• $l$ = slant height (the length of the straight line running down the outside, from apex to rim).

PICTURE. In the figure below, follow the three coloured segments. The blue $r$ lies flat. The yellow $h$ goes straight up. The red $l$ is the tilted outer edge. Notice they meet as a right triangle: $r$ along the floor, $h$ up the axis, $l$ closing the gap.

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Step 2 — Slice the cone into coins (setting up the volume)

WHAT. Cut the cone horizontally into a huge stack of very thin discs, like a tower of coins that get smaller toward the top.

WHY. We already know the area of a flat circle is $\pi (\text{radius}{)}^2$. If we can find the radius of each coin and add up all their volumes, we get the whole cone. This "add up thin slices" idea is exactly integration, and it is the one honest way to handle a shape whose width keeps changing.

PICTURE. Look at one coin at height $y$ measured up from the base. As $y$ climbs from $0$ (bottom) to $h$ (apex), the coin's radius shrinks from $r$ down to $0$.

The shrinking is a straight-line (similar-triangle) relationship — see Similar Solids. At the bottom the radius is $r$; at the top it is $0$; it falls in a straight line, so at height $y$:

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Step 3 — Add the coins: the integral that gives the volume

WHAT. Each coin has area $\pi \rho (y{)}^2$ and tiny thickness $dy$, so tiny volume $\pi \rho (y{)}^{2}dy$. Sum from $y=0$ to $y=h$.

WHY the integral and not just multiplication? For a cylinder every slice is identical, so plain area $\times$ height works. Here the slices change size, so we must add up infinitely many different ones — that summing-of-changing-things is precisely what $\int$ means.

PICTURE. The figure plots the coin area $\pi \rho (y{)}^2$ against height $y$: it starts at $\pi r^2$ and curves down to $0$. The area under that curve is the cone's volume — the yellow shaded region.

$$V={\int }_0^h\pi \rho (y{)}^{2}dy=\pi r^2{\int }_0^h{\left(1-\frac{y}{h}\right)}^{2}dy$$

Let $u=1-\frac{y}{h}$ so the bracket becomes just $u^2$; as $y$ runs $0\to h$, $u$ runs $1\to 0$, and $du=-\frac{1}{h}dy$:

$$V=\pi r^2\cdot h{\int }_0^1{u}^{2}du=\pi r^{2}h\cdot \underset{=1/3}{\underbrace{[\frac{u^3}{3}{]}_0^1}}=\boxed{\frac{1}{3}\pi r^{2}h}$$

• $\pi r^2$ — area of the widest (bottom) coin, pulled out front.
• $h$ — came from the substitution, restoring the height.
• $\frac{1}{3}$ — the entire content of the derivation: it is ${\int }_0^1{u}^{2}du$. That $u^2$ (area $\propto$ radius${}^2$) is why it is a third, not a half.

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Step 4 — Sanity check the volume against the cylinder

WHAT. Compare the cone to the cylinder that hugs it (same $r$, same $h$).

WHY. A new formula you cannot check is a formula you cannot trust. The cylinder's volume is $\pi r^{2}h$ (base circle stacked straight up, no shrinking). Our cone is one-third of that — a claim you can verify by pouring.

PICTURE. Three cone-fulls of water exactly fill one cylinder of the same base and height.

$$\frac{V_{\text{cone}}}{V_{\text{cyl}}}=\frac{\frac{1}{3}\pi r^{2}h}{\pi r^{2}h}=\frac{1}{3}✓$$

By Dimensional Analysis, both are lengths${}^3$ (a volume), so the ratio is a pure number — as it must be.

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Step 5 — Now the skin: cut the cone open and unroll it

WHAT. Leave the flat base aside. Cut the curved side straight up from rim to apex and flatten it out onto a table.

WHY unroll? We only know how to find the area of flat shapes. A curved surface must be laid flat first. This is the same trick the parent used for a cylinder (unrolling to a rectangle) — but a cone's slant means it unrolls to a pie slice (sector), not a rectangle.

PICTURE. The flattened skin is a sector of a big circle whose radius is the slant height $l$ (every straight line from apex to rim had length $l$, and those lines become the sector's straight edges). The curved arc of the sector was the base rim, so its length equals the base circumference $2\pi r$.

• Sector radius $=l$ — because each generator line (apex to rim) is length $l$ and is now a flat radius.
• Sector arc length $=2\pi r$ — the base rim did not change length when we cut and unrolled it.

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Step 6 — Area of that pie slice = the curved surface area

WHAT. Find the area of the sector.

WHY this ratio. A full circle of radius $l$ has area $\pi l^2$ and full arc (circumference) $2\pi l$. Our sector uses only part of that arc, namely $2\pi r$. Area is shared out in exactly the same proportion as arc length — take the whole circle's area and keep the same fraction.

PICTURE. The sector as a fraction of the full disc: the shaded slice is the fraction $\frac{2\pi r}{2\pi l}$ of the whole.

$$\text{CSA}=\underset{\text{fraction of full circle}}{\underbrace{\frac{2\pi r}{2\pi l}}}\times \underset{\text{full circle area}}{\underbrace{\pi l^2}}=\frac{r}{l}\cdot \pi l^2=\boxed{\pi rl}$$

• $2\pi r$ — arc we actually kept (base circumference).
• $2\pi l$ — arc of the whole circle of radius $l$.
• $\pi l^2$ — area of that whole circle.
• The $l$'s cancel to leave the clean $\pi rl$.

Add the flat base circle $\pi r^2$ for the total surface area:

$$\text{TSA}=\underset{\text{base}}{\underbrace{\pi r^2}}+\underset{\text{curved skin}}{\underbrace{\pi rl}}=\pi r(r+l)$$

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Step 7 — Edge and degenerate cases (never leave a gap)

WHAT & WHY. Check what happens when the numbers go to extremes, so no reader is ever surprised.

• Flat cone, $h\to 0$. Then $l=\sqrt{r^2+0}=r$. Volume $\frac{1}{3}\pi r^2\cdot 0=0$ (no space inside — it's a flat disc). CSA $=\pi r\cdot r=\pi r^2$, matching the base area, because the "cone" is just the disc seen from both sides.
• Needle cone, $r\to 0$. Both $V=\frac{1}{3}\pi \cdot 0\cdot h=0$ and CSA $=\pi \cdot 0\cdot l=0$ — a bare line has no volume and no skin. Correct.
• Tall thin cone, $h\gg r$. Then $l\approx h$, so CSA $\approx \pi rh$ — the slant is nearly the height, and the skin behaves almost like a cylinder wall. Sensible.
• $r=h$ (a $45^{\circ }$ cone). $l=r\sqrt{2}$, so TSA $=\pi r(r+r\sqrt{2})=\pi r^2(1+\sqrt{2})$. Nothing breaks.

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Worked example (both results at once)

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The one-picture summary

Left: shrinking coins summed by an integral $\to \frac{1}{3}\pi r^{2}h$. Right: skin unrolled to a pie slice $\to \pi rl$. One triangle ($r,h,l$) feeds both halves.

Recall Feynman retelling — say it back in plain words

I take a cone and slice it into a stack of coins. Near the bottom a coin is as wide as the base; near the top it shrinks to a dot. Because a coin's area depends on its radius squared, when I add all the coins up the total comes to exactly one-third of the full cylinder that would box the cone in — I can prove it by pouring three cones of water into one cylinder. For the outside skin, I cut the slanted side open and flatten it into a fan-shaped pizza slice. The slice's straight edges are the slant length $l$, and its curved edge is the base's rim of length $2\pi r$. Since a sector keeps the same fraction of area as it does of arc, its area is $\pi rl$. Height $h$ rules the inside; slant $l$ rules the outside; one right triangle links them by Pythagoras.

Recall Quick self-test

Why is the cone $\frac{1}{3}$ and not $\frac{1}{2}$ of the cylinder? ::: Because a slice's area shrinks with the square of its radius, and ${\int }_0^1{u}^{2}du=\frac{1}{3}$ (a linear shrink would give $\frac{1}{2}$). Which length owns the curved surface area, $h$ or $l$? ::: The slant $l$, because $\text{CSA}=\pi rl$ measures paint travelling along the tilted outside. When the cone is flattened ($h\to 0$), what does CSA become and why? ::: $\pi r^2$, because $l\to r$ and the "cone" collapses into the flat base disc.

Back to parent: Surface area and volume of all above 3D shapes.