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Calculus AB · Unit 8 · 8.9
AP Calculus · Applications of Integration
Volume with the Disc Method
Each slice sweeps a coin of radius R. Stack the coins and you have the solid.
key idea$V=\pi\int_a^b \big[R(x)\big]^{2}\,dx$
Picture the region
Spin a region about an axis and each vertical slice sweeps out a disc — a flat circular coin of radius $R$ and thickness $dx$. Its area is $\pi R^{2}$, and stacking the coins is the integral. The only real work is identifying $R$.
The region under $y=\sqrt{x}$ on $[0,4]$, revolved about the $x$-axis. The dashed rose line is the axis of revolution; the faint mirror image below it is where the region sweeps to. The amber ellipse is one disc, and its radius $R$ reaches from the axis to the curve — here simply $\sqrt{x}$.[Figure: The same region, revolved. Every slice perpendicular to the axis is a solid disc of radius R(x) — the shaded one is the pi R2 inside the integral.]
The same region, revolved. Every slice perpendicular to the axis is a solid disc of radius $R(x)$ — the shaded one is the $\pi R^2$ inside the integral.Setting up the integral
Building the disc integral
- Identify the axis of revolution and draw it. Every radius will be measured from this line.
- Confirm the region actually touches the axis along the whole interval. If there is a gap, the slice is a ring rather than a disc and this method does not apply.
- Draw one slice perpendicular to the axis. About the $x$-axis that means a vertical slice and $dx$; about the $y$-axis, a horizontal slice and $dy$.
- Write $R$ as the distance from the axis to the outer boundary. For the $x$-axis, $R=f(x)$; for the $y$-axis, rewrite the curve as $x=g(y)$ and take $R=g(y)$.
- Integrate $\pi R^{2}$ over the limits, and square before integrating — never after.
four commitments made at set-up
- The slice must be perpendicular to the axis of revolution. That single choice fixes the differential.
- $R$ is a distance from the axis, so it is always positive, even where the curve is below the axis.
- Revolving about the $y$-axis means solving for $x$ and integrating $dy$ — both halves switch together.
- $\pi$ is a constant: pull it outside the integral and keep it out of the antidifferentiation.
Worked examples
EX 1
About the $x$-axis
$y=\sqrt{x}$ on $[0,4]$The region under $y=\sqrt{x}$ from $x=0$ to $x=4$ is revolved about the $x$-axis. Find the volume.
1
Identify the radius
The slice runs from the axis to the curve, so $R=\sqrt{x}$.
2
Square it
$R^{2}=x$ — the square root disappears, which is why this method suits radical boundaries.
3
Set up
$V=\pi\displaystyle\int_{0}^{4}x\,dx$.
4
Integrate
$\pi\left[\dfrac{x^{2}}{2}\right]_{0}^{4}$.
5
Evaluate
$\pi\cdot8=8\pi$.
EX 2
About the $y$-axis
solve for $x$ firstThe region bounded by $y=x^{2}$, $y=4$ and the $y$-axis is revolved about the $y$-axis. Find the volume.
1
Choose the slice
Perpendicular to the $y$-axis means horizontal slices and a $dy$ integral.
2
Rewrite the boundary
$y=x^{2}$ becomes $x=\sqrt{y}$ for $x\ge0$.
3
Identify the radius
$R=\sqrt{y}$, measured from the $y$-axis out to the curve.
4
Set up
$V=\pi\displaystyle\int_{0}^{4}\left(\sqrt{y}\right)^{2}dy=\pi\int_{0}^{4}y\,dy$.
5
Evaluate
$\pi\left[\dfrac{y^{2}}{2}\right]_{0}^{4}=8\pi$.
EX 3
A region below the axis
distance, not valueThe region between $y=-x^{2}$ and the $x$-axis on $[0,2]$ is revolved about the $x$-axis. Find the volume.
1
Note where the region sits
Entirely below the axis, since $-x^{2}\le0$.
2
Write the radius as a distance
$R=\left\lvert-x^{2}\right\rvert=x^{2}$.
3
Observe that squaring settles it
$R^{2}=\left(-x^{2}\right)^{2}=x^{4}$ either way — the sign cannot survive.
4
Set up and integrate
$V=\pi\displaystyle\int_{0}^{2}x^{4}dx=\pi\left[\dfrac{x^{5}}{5}\right]_{0}^{2}$.
5
Evaluate
$\dfrac{32\pi}{5}$. Being below the axis changes nothing, because a radius is a distance.
EX 4
Recovering a known formula
the coneRevolve the region under $y=\dfrac{r}{h}x$ from $x=0$ to $x=h$ about the $x$-axis, and identify the solid.
1
Identify the radius
$R=\dfrac{r}{h}x$, growing linearly from $0$ to $r$.
2
Set up
$V=\pi\displaystyle\int_{0}^{h}\dfrac{r^{2}}{h^{2}}x^{2}dx$.
3
Pull out the constants
$\dfrac{\pi r^{2}}{h^{2}}\displaystyle\int_{0}^{h}x^{2}dx=\dfrac{\pi r^{2}}{h^{2}}\cdot\dfrac{h^{3}}{3}$.
4
Simplify
$V=\dfrac{1}{3}\pi r^{2}h$.
5
Recognise it
The volume of a cone of radius $r$ and height $h$ — the standard formula, derived rather than memorised.
Watch out
| Axis of revolution | Slice direction | Radius | Integrate |
|---|
| the $x$-axis | vertical | $R=f(x)$ | $\pi\displaystyle\int f^{2}dx$ |
| the $y$-axis | horizontal | $R=g(y)$ | $\pi\displaystyle\int g^{2}dy$ |
| a horizontal line $y=k$ | vertical | $R=\lvert f(x)-k\rvert$ | $\pi\displaystyle\int(f-k)^{2}dx$ |
| a vertical line $x=k$ | horizontal | $R=\lvert g(y)-k\rvert$ | $\pi\displaystyle\int(g-k)^{2}dy$ |
Rows 1 and 2 are this topic; rows 3 and 4 are the same idea with the radius measured from a shifted line. The structure never changes — only what you subtract inside the square.
Square first, then integrate
$\int f^{2}$ is not $\left(\int f\right)^{2}$. Expand the square completely before antidifferentiating.
The slice is perpendicular to the axis
Revolving about the $y$-axis with a $dx$ integral is a set-up error no amount of correct algebra can rescue.
Discs require the region to touch the axis
If there is a gap between the region and the axis, each slice is a ring and a second radius is needed.
Keep $\pi$ outside
It is a constant factor. Carrying it through the antidifferentiation invites dropping it — and a volume without $\pi$ is visibly wrong.