Volumes of revolution about the x-axis and y-axis, volumes generated by the region between two curves, and parametric and improper cases, using integration of pi y squared or pi x squared.

What this dot point is asking

OCR wants you to find the volume of a solid of revolution: rotate a region about the $x$-axis using $V = \pi\int y^2\,dx$, rotate about the $y$-axis using $V = \pi\int x^2\,dy$, find the volume of the solid between two curves as a difference of two such integrals, and handle parametric curves and improper (infinite) cases by the same method.

Rotation about the x-axis

Imagine slicing the solid into thin discs perpendicular to the $x$-axis. Each disc has radius $y$, so cross-sectional area $\pi y^2$ and volume $\pi y^2\,dx$; summing them by integration gives the total volume.

Rotation about the y-axis

Rotation about the $y$-axis uses horizontal discs of radius $x$. You must express $x^2$ as a function of $y$ and integrate between $y$-limits.

The solid between two curves

When the region lies between two curves, rotation produces a solid with a hole (a washer at each slice). The volume is the outer solid minus the inner solid, so you subtract the squares of the two functions inside one integral.

Parametric and improper cases

For a parametric curve $x = x(t)$, $y = y(t)$, rotation about the $x$-axis gives $V = \pi\displaystyle\int y^2\,\dfrac{dx}{dt}\,dt$, with the limits converted from $x$-values to the corresponding values of $t$. The trick is to leave $y$ in terms of $t$, substitute $dx = \dfrac{dx}{dt}\,dt$, and integrate with respect to $t$; converting back to $x$ first usually creates a messier integral. If the region extends to infinity, for example rotating $y = e^{-x}$ on $0 \le x < \infty$, the result is an improper integral, which you evaluate by replacing the infinite limit with a variable and taking a limit exactly as in the improper-integrals topic. Such a solid can have a finite volume even though it is infinitely long, because the squared radius decays fast enough for the integral to converge.

Volumes of revolution combine integration technique with geometric visualisation and reuse the improper-integral idea whenever the region is unbounded.

Try this

Q1. Find the volume when $y = 2$ (a horizontal line), between $x = 0$ and $x = 3$, is rotated about the $x$-axis. [2 marks]

• Cue. $V = \pi\int_0^3 2^2\,dx = \pi[4x]_0^3 = 12\pi$ (a cylinder of radius $2$, length $3$).

Q2. Write the integral for the volume when $y = x^2$, between $y = 0$ and $y = 4$, is rotated about the $y$-axis. [2 marks]

• Cue. $x^2 = y$, so $V = \pi\int_0^4 y\,dy = 8\pi$.