5.17 Areas and volumes of revolution

Integration was first used to find the area between a curve and the $x$-axis (5.11). The same idea, cutting a shape into thin strips and adding them up, also finds areas measured against the $y$-axis, and volumes of solids. When a flat region is rotated one full turn about a line, it sweeps out a solid whose cross-sections are all circles. Many manufactured objects are made this way: a potter's wheel or a lathe turns material about a fixed axis, so vases, bowls, wine glasses, bottles and machine parts are solids of revolution.

The volume of such a solid depends only on the curve that forms its profile. Slicing the solid into thin discs and adding their volumes with an integral gives the volume exactly, which allows a designer to calculate how much a container holds, how deep a given amount of liquid will be, or how much material a part needs.

Area between a curve and the y-axis

Integrating with respect to y

The area between a curve and the $x$-axis is found by cutting the region into thin vertical strips. Each strip has height $y$ and width $\delta x$, so its area is approximately $y\,\delta x$, and adding the strips and letting $\delta x \to 0$

A region bounded by a curve and the $y$-axis is cut into thin horizontal strips instead. Each strip has length $x$, the distance from the $y$-axis to the curve, and thickness $\delta y$, so its area is approximately $x\,\delta y$. Adding the strips from $y = c$

The roles of $x$ and $y$ are simply exchanged. The curve must be written as $x = g(y)$, with $x$ as a function of $y$, and the limits $c$

To rearrange $y = f(x)$ into $x = g(y)$, make $x$ the subject, as when finding an inverse function (2.2). For example, $y = x^3$

The check shows that integrating with respect to $y$ is not a new kind of area. It is the same area found by slicing in a different direction, and the choice of direction is made to keep the calculation simple.

Curves on both sides of the y-axis

For a region to the left of the $y$-axis, $x$ is negative, so $\int_c^d x\,dy$ is negative. This matches the behaviour of $\int_a^b y\,dx$

Choosing between dx and dy

Integrating with respect to $y$ is often the better choice for a region bounded on the left and right by two curves. Each horizontal strip then runs from the left curve to the right curve, and its length is $x_{\text{right}} - x_{\text{left}}$, so

where $c$ and $d$ are the $y$-coordinates of the points where the curves meet. With vertical strips, the same region might have a different top or bottom boundary in different places, so the area would have to be split into several integrals.

Volumes of revolution

Solids of revolution

When a region in the plane turns through one full revolution ($2\pi$ radians) about a fixed line, it sweeps out a solid. Every point of the region moves in a circle centred on the line, so every cross-section of the solid perpendicular to the line is a circle (or a ring, if the region does not touch the line).

Familiar solids arise in this way. Rotating a rectangle about one of its sides gives a cylinder; rotating a right-angled triangle about one of its shorter sides gives a cone; rotating a semicircle about its diameter gives a sphere. The formulas for their volumes (3.1) can all be derived by integration, and the same method finds volumes of solids that have no simple formula.

Rotating about the x-axis: the disc method

Consider the region under the curve $y = f(x)$, above the $x$-axis, between $x = a$ and $x = b$. Rotating it about the $x$

Adding the volumes of all the discs from $x = a$ to $x = b$ approximates the volume of the solid. As $\delta x \to 0$ the approximation becomes exact, and the sum becomes an integral, in the same way as the area under a curve in 5.5 and 5.11.

Squaring $y$ before integrating is the key step. The integrand is $\pi y^2$, the area of a circular cross-section, so a volume integral adds up cross-sectional areas, in the same way that an area integral adds up lengths.

Cones and spheres

Two of the volume formulas in 3.1 can now be proved. A cone is formed by rotating a straight line through the origin, and a sphere by rotating a semicircle.

The radius of the cone grows in proportion to $x$, so the area of each circular slice grows in proportion to $x^2$. Integrating $x^2$ produces the factor $\frac{1}{3}$

Volumes involving other functions

Any function that can be squared and integrated can be rotated. Trigonometric functions often need an identity before integrating, and products such as $(\ln x)^2$ may need integration by parts (5.16).

When a region lies below the $x$-axis, $y$ is negative but $y^2$ is positive, so $\int_a^b \pi y^2\,dx$

Rotating about the y-axis

Rotating a region about the $y$-axis works in the same way with $x$ and $y$ exchanged. The slices are now horizontal discs of thickness $\delta y$, and the radius of each disc is the horizontal distance $x$ from the $y$

The bowl formed by rotating $y = x^2$ about the $y$-axis has the same volume as the solid formed by rotating $y = \sqrt{x}$

Regions between two curves

The washer method

When the region being rotated does not touch the axis of rotation along its whole length, the solid has a hole through it. A thin slice of such a solid is a flat ring, called a washer, with an outer radius $R$ and an inner radius $r$. Its volume is the volume of the outer disc minus the volume of the hole:

For a region between two curves $y = f(x)$ (the outer one) and $y = g(x)$ (the inner one), with $f(x) \geq g(x) \geq 0$

Washers about the y-axis

About the $y$-axis the washers are horizontal, the radii are horizontal distances, and both curves must be written as $x$ in terms of $y$:

Regions that need splitting

Sometimes the inner or outer boundary changes part of the way along the axis. The solid is then split where the boundary changes, and the volumes of the parts are added. This happens often when a region under a curve $y = f(x)$, with $x$ between two positive values, is rotated about the $y$-axis.

Modelling real objects

A container with circular horizontal cross-sections can be modelled by rotating a curve about a vertical axis. Placing the axis of the container along the $y$-axis, with $y$ the height above the base and $x$ the inner radius at that height, the volume of liquid up to a depth $h$ is

The capacity is $V$ at the full height. The depth for a given volume is the solution of $V(h) =$ the volume, which usually needs a GDC.

Exam-style questions on areas and volumes of revolution

Self-check

• To do:

  I can rearrange $y = f(x)$ into $x = g(y)$ and find the area between a curve and the $y$-axis using $\int_c^d |x|\,dy$