Calculate the area under a curve, the area between two curves, and the volume of a solid of revolution about the $x$ or $y$ axis

What this dot point is asking

NESA wants you to interpret a definite integral as an area, extend that to the area between two curves, and apply the disk method to compute the volume of a solid formed by rotating a region about the $x$ or $y$ axis.

The answer

If $f(x) \geq 0$ on $[a, b]$, then the area between $y = f(x)$ and the $x$-axis is

If $f(x) < 0$ on part of the interval, the integral subtracts that part. To get the geometric area (always positive), split the integral at zeros of $f$ and take absolute values:

Area between two curves

If $f(x) \geq g(x)$ on $[a, b]$, the area between them is

Find the intersection points by solving $f(x) = g(x)$ and use them as your limits. If the curves cross inside the interval, split the integral at each crossing and use the correct ordering of "upper minus lower" on each piece.

The reason "upper minus lower" works regardless of the axis is worth understanding, because it removes the need to worry about whether the curves are above or below the $x$-axis. The vertical gap between the two curves at a given $x$ is $f(x) - g(x)$ whether both are positive, both negative, or straddling the axis: any constant you might add to shift both curves up cancels in the difference. So you never split at the $x$-axis for an area-between-curves problem, only at the points where the two curves themselves cross.

Area between a curve and the $y$-axis

If the boundary is described as $x = h(y)$, integrate with respect to $y$:

Volume of revolution about the $x$-axis

Rotating the region under $y = f(x)$ between $x = a$ and $x = b$ about the $x$-axis produces a solid whose cross sections perpendicular to the axis are disks of radius $f(x)$. The volume is

The $[f(x)]^2$ is the part students drop. It is there because the cross section is a circle of radius $f(x)$, and a circle's area is $\pi r^2$, so the radius gets squared before integrating, not after. This is also why you cannot pull the $\pi$ inside or skip it: every cross section is a full disk, and $\pi$ is the area of a unit-radius one.

The volume of revolution, stage by stage

The disk method is best seen as a film: a flat region spins about an axis and sweeps out a solid, which you then slice back into disks to add up. The four stages below build exactly that picture for rotation about the $x$-axis.

Stage 1, draw the region. Shade the region bounded by $y = f(x)$, the $x$-axis and the lines $x = a$ and $x = b$. This flat region is what gets rotated; everything else follows from it.

Stage 2, take a representative disk. Picture a thin vertical strip of the region at a typical $x$. When the region spins, that strip sweeps out a thin disk whose radius is the height of the curve, $r = f(x)$, and whose thickness is $dx$. Its circular face is the ellipse straddling the axis.

Stage 3, sweep the whole region. A full turn about the $x$-axis sends every strip around, sweeping the region into a solid. Its outline is the curve above the axis and its mirror image below; the far end is a circular cap of radius $f(b)$.

Stage 4, sum the disks. Fill the solid with these thin disks. One disk has volume $\pi r^2 \, dx = \pi [f(x)]^2 \, dx$ (area of a circle times thickness), and adding them all from $a$ to $b$ turns the sum into an integral:

Volume of revolution about the $y$-axis

Rotating the region between $x = h(y)$ and the $y$-axis, between $y = c$ and $y = d$, about the $y$-axis gives

You must invert the relation to express $x$ as a function of $y$ before integrating.

Volume between two curves

If a region is bounded by $y = f(x)$ above and $y = g(x)$ below, rotating about the $x$-axis gives a washer with outer radius $f(x)$ and inner radius $g(x)$:

Note that this is not $\pi \int (f - g)^2$. The reason is geometric: each cross section is an annulus (a washer), whose area is the big circle minus the small hole, $\pi R^2 - \pi r^2$. That is $\pi(R^2 - r^2)$, the difference of the squared radii. Squaring the gap $(f - g)$ would measure the area of a disk whose radius is the gap, which is a completely different solid.

How exam questions ask about areas and volumes

The wording pins down both the method and the variable you integrate in:

• "Find the area bounded by / enclosed by the curve and the $x$-axis." A single curve against the axis: integrate $f(x)$, but if the curve dips below the axis, split at its roots and take magnitudes so the area is positive.
• "Find the area between / enclosed by the two curves." Solve $f(x) = g(x)$ for the limits, then integrate upper minus lower. Sketch first so you know which curve is on top.
• "The region is rotated about the $x$-axis." Disk method in $x$: $V = \pi \int [f(x)]^2\,dx$, with $x$-limits.
• "The region is rotated about the $y$-axis." Disk method in $y$: rearrange to $x = h(y)$ first, then $V = \pi \int [h(y)]^2\,dy$, with $y$-limits. The single most common error here is forgetting to invert and convert the limits to $y$-values.
• "Find the volume of the solid between $y = f(x)$ and $y = g(x)$ when rotated about the $x$-axis." Washer method: $V = \pi \int ([f]^2 - [g]^2)\,dx$, outer radius squared minus inner radius squared.
• "Give your answer in exact form" or no rounding instruction. Leave $\pi$, surds and fractions in place; do not reach for the calculator.

The axis of rotation decides the variable of integration every time: rotate about the $x$-axis and you integrate in $x$; rotate about the $y$-axis and you integrate in $y$. Get that pairing wrong and the whole integral is set up incorrectly.