Visual walkthrough — 3D shapes — cube, cuboid, cylinder, cone, sphere, prism, pyramid

Before we start, three plain-word promises about the letters we will use:

• $r$ = the radius of the ball: the distance from its dead centre to its skin.
• $V$ = volume: how much "stuff" (water, sand) fits inside.
• A curly $\int$ (an "integral") will appear later — we build it from scratch, so ignore it for now.

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Step 1 — What is a sphere, really?

WHAT. A sphere is the set of all points that sit exactly the same distance $r$ from one fixed centre point. Not "roughly round" — perfectly round. Every point on the skin is $r$ away from the middle.

WHY start here. If we do not pin down what makes a point belong to the sphere, we cannot describe any slice of it. The rule "distance from centre $=r$" is the single fact the whole derivation squeezes.

PICTURE. Look at the figure. The green dot is the centre. The blue skin is every point at distance $r$. The orange arrow is $r$ — pick any direction, its length is always the same.

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Step 2 — Set up a ruler through the middle

WHAT. We slide a vertical ruler (call its readings $y$) through the centre. The centre reads $y=0$. The top of the ball is at $y=+r$, the bottom at $y=-r$.

WHY this tool. To slice the ball into thin horizontal coins, we need an address for each height. The number $y$ is that address: "the coin sitting at height $y$." Everything downstream is written in terms of $y$.

PICTURE. The vertical orange axis is the ruler. $y$ runs from $-r$ (bottom) up through $0$ (centre) to $+r$ (top). One horizontal blue line marks the coin at a chosen height $y$.

• $y=0$ ::: the height of the centre (widest part of the ball).
• $y=+r$ ::: the very top point (a coin of zero size).
• $y=-r$ ::: the very bottom point (also zero size).

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Step 3 — How wide is the coin at height $y$?

WHAT. Take the flat coin at height $y$. It is a circle. Its own radius — call it $x$ — is not $r$; it is smaller, and shrinks to $0$ as $y$ approaches the poles. We find $x$ using a right triangle.

WHY a triangle / why Pythagorean Theorem. Draw the line from the centre to a point on the coin's rim. That line has length $r$ (it lands on the sphere's skin!). It is the hypotenuse. Its vertical leg is $y$ (how high we climbed). Its horizontal leg is $x$ (the coin's radius). A right angle sits between the vertical and horizontal legs. The Pythagorean theorem is the tool that links the three sides of a right triangle — that is exactly the relationship we need.

$\underset{\text{coin radius, squared}}{\underbrace{x^2}}+\underset{\text{height, squared}}{\underbrace{y^2}}=\underset{\text{sphere radius, squared}}{\underbrace{r^2}}$

Solve for the coin radius:

$x^2=r^2-y^2\Rightarrow x=\sqrt{r^2-y^2}$

PICTURE. The red hypotenuse is $r$ (fixed). The orange leg is $y$. The blue leg is $x$, the coin's radius. As the blue horizontal line slides up, the red length stays the same but $y$ grows, so $x$ must shrink — the coins get smaller near the top.

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Step 4 — Area of one coin

WHAT. Each coin is a circle of radius $x$. The area of a circle is $\pi x^2$ (from Circle Geometry). We already know $x^2=r^2-y^2$, so we substitute:

$A(y)=\pi x^2=\pi \left(\underset{\text{= }{x}^2\text{ from Step 3}}{\underbrace{r^2-y^2}}\right)$

WHY write it as $A(y)$. The notation $A(y)$ just means "the area depends on the height $y$." Feed in a height, get back that coin's area. At $y=0$: $A=\pi r^2$ (biggest coin). At $y=\pm r$: $A=\pi (r^2-r^2)=0$ (the poles are points, zero area). The formula already handles both degenerate ends automatically — a good sign.

PICTURE. A blue disc of radius $x$ sits at height $y$; its label reads $A(y)=\pi (r^2-y^2)$. A tiny arrow shows its thickness $\mathrm{d}y$ — the coin is thin, like a real coin.

• $\pi$ ::: the circle constant, appears because each slice is a circle.
• $A(y)$ ::: the flat area of the coin sitting at height $y$.

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Step 5 — Volume of one thin coin, then add them all

WHAT. A coin is a very short cylinder: area $\times$ thickness. Give it thickness $\mathrm{d}y$ (a tiny height). Its sliver of volume is

$\mathrm{d}V=\underset{\text{coin’s flat area}}{\underbrace{A(y)}}\times \underset{\text{coin’s tiny thickness}}{\underbrace{\mathrm{d}y}}=\pi (r^2-y^2)\mathrm{d}y$

WHY the integral $\int$. We must add up infinitely many coins from bottom ($y=-r$) to top ($y=+r$). Ordinary $+$ can't add infinitely many infinitely thin things. The symbol ${\int }_{-r}^r$ is exactly the machine built to do that — "sum every sliver as the thickness shrinks to zero." That is why Integration enters here and not simple multiplication: the coins keep changing size, so no single multiplication covers them all.

$V={\int }_{-r}^r\pi (r^2-y^2)\mathrm{d}y$

PICTURE. A stack of blue coins fills the sphere's outline: fat in the middle, thin at the poles. The arrow shows summing from $-r$ up to $+r$.

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Step 6 — Do the sum (evaluate the integral)

WHAT. Pull the constant $\pi$ out front and integrate term by term. The rule we use: the running-total of $r^2$ (a constant in $y$) is $r^{2}y$, and the running-total of $y^2$ is $\frac{y^3}{3}$.

$V=\pi {\left[\underset{\text{from }{r}^2}{\underbrace{r^{2}y}}-\underset{\text{from }{y}^2}{\underbrace{\frac{y^3}{3}}}\right]}_{-r}^r$

WHY. "Evaluate from $-r$ to $r$" means: plug in the top value, then subtract the bottom value. This is how a definite integral turns into an actual number.

Top ($y=r$): $r^2(r)-\frac{r^3}{3}=r^3-\frac{r^3}{3}=\frac{2r^3}{3}$

Bottom ($y=-r$): $r^2(-r)-\frac{(-r{)}^3}{3}=-r^3+\frac{r^3}{3}=-\frac{2r^3}{3}$

Subtract bottom from top:

$V=\pi \left(\frac{2r^3}{3}-\left(-\frac{2r^3}{3}\right)\right)=\pi \cdot \frac{4r^3}{3}=\boxed{\frac{4}{3}\pi r^3}$

PICTURE. A number line for $y$ marks the top contribution $+\frac{2r^3}{3}$ and the bottom contribution $-\frac{2r^3}{3}$; the gap between them, $\frac{4r^3}{3}$, is shaded — that gap, times $\pi$, is the volume.

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Step 7 — Sanity checks: the edge cases

WHAT / WHY. A formula you trust must survive its extreme inputs.

• Zero radius $r=0$: $V=\frac{4}{3}\pi (0{)}^3=0$. A ball of no size holds nothing. ✓
• Doubling the radius $r\to 2r$: $V=\frac{4}{3}\pi (2r{)}^3=\frac{4}{3}\pi \cdot 8r^3=8V_{\text{old}}$. Twice as wide holds eight times as much — because volume grows with the cube of size (see Units and Measurement). ✓
• Compare to its cylinder: a cylinder that just wraps the ball has radius $r$, height $2r$, so $V_{\text{cyl}}=\pi r^2(2r)=2\pi r^3$. Our sphere is $\frac{4}{3}\pi r^3$, which is exactly $\frac{2}{3}$ of that cylinder — Archimedes' famous ratio.

PICTURE. A small ball and a ball of double radius, side by side, the big one broken into 8 small-ball-sized chunks to show the $\times 8$ jump.

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The one-picture summary

Below: the whole story in one frame — triangle gives the coin radius $x=\sqrt{r^2-y^2}$, the coin's area is $\pi (r^2-y^2)$, stacking the coins from $-r$ to $r$ is the integral, and the total is $\frac{4}{3}\pi r^3$.

Recall Feynman retelling — say it back in plain words

I want to know how much water fits in a ball. I slice the ball into thin coins, stacked from bottom to top. Each coin is a circle, but its size depends on how high up it sits: right through the middle it's as wide as the ball, and near the top and bottom it shrinks to a dot. A right triangle tells me a coin at height $y$ has radius $\sqrt{r^2-y^2}$ — because the line from the centre to the coin's rim is always length $r$, the ball's radius. So the coin's area is $\pi (r^2-y^2)$. I can't just multiply, because every coin is a different size, so I add up all the coins with an integral from the bottom ($-r$) to the top ($+r$). Grinding that sum out gives $\frac{4}{3}\pi r^3$. The $\pi$ is because the coins are circles; the $r^3$ is because volume is a three-dimensional (length-cubed) thing; and the funny $\frac{4}{3}$ is the leftover from the coins fading to nothing at the poles.

Recall Quick self-test

Why is the coin at $y=0$ the widest? ::: Because $x=\sqrt{r^2-0}=r$ there; any other $y$ makes $x$ smaller. Why use an integral instead of multiplying area by height? ::: The coins change size with height, so no single area covers the whole ball — we must sum infinitely many changing slices. If the radius triples, the volume multiplies by? ::: $3^3=27$.

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See also

• Pythagorean Theorem — gives the coin radius.
• Circle Geometry — gives each coin's area $\pi x^2$.
• Integration — the tool that adds the coins.
• Area of2D Shapes · Units and Measurement · Real-World Applications