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

Area under a curve

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.

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

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 $\pi \int (f - g)^2$.

Worked example: area between curves

Find the area between $y = x^3$ and $y = x$ for $x \in [0, 1]$.

On $[0, 1]$, $x \geq x^3$, so

$A = \int_0^1 (x - x^3) \, dx = \left[ \frac{x^2}{2} - \frac{x^4}{4} \right]_0^1 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$ square units.

Worked example: volume about the $y$-axis

The region bounded by $y = x^2$, $y = 4$ and the $y$-axis (first quadrant) is rotated about the $y$-axis. Find the volume.

Solve for $x$ in terms of $y$: $x = \sqrt{y}$, so $x^2 = y$.

$V = \pi \int_0^4 y \, dy = \pi \left[ \frac{y^2}{2} \right]_0^4 = 8 \pi$ cubic units.

Common traps

Forgetting the absolute value. A definite integral can be negative if the curve dips below the axis. Area is non-negative, so split or take absolute values.

Squaring the difference instead of differencing the squares. Volume of a washer uses $[f(x)]^2 - [g(x)]^2$, not $(f(x) - g(x))^2$.

Integrating with respect to the wrong variable. Rotation about the $x$-axis pairs with $\int f(x)^2 \, dx$. Rotation about the $y$-axis pairs with $\int h(y)^2 \, dy$.

Missing intersections. If the curves cross inside your interval, the upper and lower roles swap. Split the integral and re-test which curve is on top in each subinterval.

Dropping $\pi$. A volume of revolution always carries a factor of $\pi$. Forgetting it costs marks.

In one sentence

Areas are definite integrals of "upper minus lower" and volumes of revolution are $\pi \int r^2$, with $r$ being the perpendicular distance from the rotation axis to the boundary.