9C - Volumes of revolution

9C - Volumes of revolution

What solid appears when a flat curved region spins around an axis? You will build its volume from thin circular discs, identify the correct radius, and integrate with respect to the coordinate running along the axis of rotation.

Building volume from discs

A narrow strip joining a curve to an axis sweeps out a thin circular disc when the region makes one complete revolution.

Imagine the rectangle 0≤x≤50\le x\le5, 0≤y≤20\le y\le2 rotating about the xx-axis. Every cross-section perpendicular to that axis is a circle of radius 22. Its area is 4π4\pi; stacking those cross-sections over length 55 gives a cylinder of volume 20π20\pi.

For a curved boundary the radius changes along the solid. A very thin slice can be approximated by a disc whose radius is the curve's distance from the axis at that position. Adding increasingly thin discs leads to an integral. This explains the formula without requiring a formal limiting proof.

Volume from circular cross-sections

A thin disc of radius rr and thickness Δx\Delta x has approximate volume

πr2Δx.\pi r^2\Delta x.

Integrating the circular cross-sectional area along the solid gives its volume. Radius squared is essential: an area integral of the radius alone does not give a volume.

[DIAGRAM: asset_name: Radius and thickness in the two orientations; asset_slug: volume-discs; caption: The radius is perpendicular to the rotation axis, while the thickness runs along it; description: Two coordinate sketches show a shaded region under y equals square root of x rotated about the x-axis and a shaded region to the left of x equals square root of y rotated about the y-axis. A vertical strip in the first is labelled radius y and thickness dx. A horizontal strip in the second is labelled radius x and thickness dy. Circular end-view sketches show the corresponding radius and disc cross-section.]
The radius is perpendicular to the rotation axis, while the thickness runs along it

For rotation about the xx-axis, the distance from the axis is ∣y∣|y|, so its square is y2y^2, and thickness runs in the xx-direction. For rotation about the yy-axis the squared radius is x2x^2, and thickness runs in the yy-direction. A negative coordinate does not make a negative radius or a negative volume.

These disc formulae apply when each perpendicular cross-section is a full disc: the specified region runs from the axis to its outer boundary. Identify the whole region before using them. We will use rotations about the coordinate axes.

The dimensions provide a useful check: circular area has square units, thickness has length units, and their product has cubic units.

Revolution about the x-axis

Rotation about the horizontal axis turns the vertical height of a region into the radius of each disc.

For the region between y=f(x)y=f(x), the xx-axis and x=a,bx=a,b, each cross-section has area π[f(x)]2\pi[f(x)]^2. Use the stated boundaries, finding an intercept if it supplies one of them. Square the entire expression for yy before integrating; keep the factor π\pi.

Rotation about the x-axis

For full circular cross-sections and a<ba<b,

V=π∫aby2 dx.V=\pi\int_a^b y^2\,\mathrm{d}x.

The coordinate-axis volume formulae must be learnt; they are not on the examination formulae sheet.

Worked example 1: squaring a linear radius

Question. The region bounded by y=x+1y=x+1, the xx-axis and x=0,2x=0,2 rotates once about the xx-axis. Find the volume formed.

Plan. Use radius x+1x+1, square the bracket and integrate from 00 to 22.

V=π∫02(x+1)2 dx=π∫02(x2+2x+1) dx=π[x33+x2+x]02=π(83+4+2)=26π3\begin{aligned} V&=\pi\int_0^2(x+1)^2\,\mathrm{d}x =\pi\int_0^2(x^2+2x+1)\,\mathrm{d}x\\ &=\pi\left[\frac{x^3}{3}+x^2+x\right]_0^2 =\pi\left(\frac83+4+2\right)=\frac{26\pi}{3} \end{aligned}

Answer. 26π3\frac{26\pi}{3} cubic units.

Check. The radii lie between 11 and 33, so the volume lies between those of cylinders of length 22 and radii 1,31,3: 2π<V<18π2\pi<V<18\pi. Also, the antiderivative differentiates to (x+1)2(x+1)^2, including the middle term.

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