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Pure · Integration

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Volumes of revolution

Rotating a region about the x- or y-axis to make a solid, and finding its volume with V = π∫y² dx or V = π∫x² dy — squaring before integrating, matching the limits to the axis of rotation, and subtracting one volume from another.

Pure · Integration

Volumes of revolution

Rotating a region about the x- or y-axis to make a solid, and finding its volume with V = π∫y² dx or V = π∫x² dy — squaring before integrating, matching the limits to the axis of rotation, and subtracting one volume from another.

Why it works

Spin the region under a curve about an axis and it sweeps out a solid. To find its volume, slice the solid into thin discs perpendicular to the axis.

Rotating about the xx-axis, a slice at position xx of thickness δx\delta x is a disc whose radius is the height of the curve, yy. Its volume is πy2 δx\pi y^2\,\delta x. Adding the slices and letting δx→0\delta x \to 0: V=π∫aby2 dx.V = \pi\int_a^b y^2\,\mathrm{d}x.

Rotating about the yy-axis, the discs are horizontal, the radius is now the horizontal distance xx, and the thickness is δy\delta y: V=π∫cdx2 dy.V = \pi\int_c^d x^2\,\mathrm{d}y.

Square first, then integrate. ∫y2 dx\int y^2\,\mathrm{d}x is not the square of ∫y dx\int y\,\mathrm{d}x. Expand or simplify y2y^2 into something you can integrate term by term before doing anything else. For y=x(4−x)y = x(4 - x), you must expand y2=x2(4−x)2=16x2−8x3+x4y^2 = x^2(4-x)^2 = 16x^2 - 8x^3 + x^4 — integrating first and squaring after gives a completely different (wrong) number.

Everything must match the axis. This is the step that decides the question:
  • About the xx-axis: integrand in terms of xx, dx\mathrm{d}x, and **limits are
xx-values**.
  • About the yy-axis: integrand in terms of yy, dy\mathrm{d}y, and **limits are
yy-values.

So for a yy-axis rotation you must rearrange the equation to get x2x^2 in terms of yy — and if the question gives you xx-limits, convert them into yy-limits through the equation of the curve first.

Keep π\pi outside. It is a constant, so factor it out and integrate what is left. Answers are normally left as an exact multiple of π\pi, such as 32π5\frac{32\pi}{5}; only give a decimal if the question asks for one.

Two curves. For the region between two curves rotated about the xx-axis, rotate each boundary and subtract the inner solid from the outer: V=π∫ab(youter 2−yinner 2)dx.V = \pi\int_a^b \left(y_{\text{outer}}^{\,2} - y_{\text{inner}}^{\,2}\right)\mathrm{d}x. Subtract the squares, not the curves — ∫(y12−y22)\int (y_1^2 - y_2^2) is not ∫(y1−y2)2\int (y_1 - y_2)^2.

A sanity check you can always run.** A straight line through the origin generates a cone, and you know its volume independently: V=13πr2hV = \tfrac13\pi r^2 h. If your method reproduces that, it is sound.