Area and volume

This unit looks like a list of formulas and is not. Every problem in it is the same three steps:

1. Slice the region into pieces of thickness $dx$ or $dy$.
2. Write one slice’s contribution — its area or volume.
3. Integrate across the range the slices cover.

Memorising four formulas is harder than learning that, and it fails on the fifth problem.

Area between curves

A slice is a thin rectangle of height “top minus bottom” and width $dx$:

$$A = \int_a^b \big(f(x) - g(x)\big)\,dx$$

For the region between $y = x$ and $y = x^2$ from $0$ to $1$, the line is above the parabola throughout:

So the area is $\tfrac16$. The verified lines are the antiderivative and the evaluation, which is the whole calculation.

Volume by discs

Rotate a region about an axis and each slice sweeps out a disc of radius equal to the function’s value:

$$V = \pi\int_a^b \big[f(x)\big]^2\,dx$$

Rotating $y = \sqrt{x}$ from $0$ to $4$ about the $x$-axis:

The radius is $\sqrt x$, so the radius squared is just $x$, and the volume is $8\pi$.

That $\pi r^2$ is the area of a circle — the disc formula is not a new fact, it is step 2 of the procedure with a circular slice.

Volume by washers

When the solid has a hole, the slice is an annulus: subtract the inner disc from the outer.

$$V = \pi\int_a^b \left(R_{\text{outer}}^2 - R_{\text{inner}}^2\right)dx$$

That is the region between $y = x$ and $y = x^2$ from $0$ to $1$ rotated about the $x$-axis: outer radius $x$, inner radius $x^2$, volume $\tfrac{2\pi}{15}$.

Known cross-sections

The general case, and the one that shows the formulas were never the point. If the slice is a square, a semicircle, or an equilateral triangle rather than a disc, only step 2 changes:

with $s$ the distance between the bounding curves at that $x$.

Squares on the region under $y = x$ from $0$ to $1$ give $\int_0^1 x^2 dx = \tfrac13$; equilateral triangles on the same base give $\tfrac{\sqrt3}{12}$. Same slicing, same limits, different slice area.

Choosing the variable

Slice perpendicular to the axis of rotation, and in the direction that avoids splitting the region.

• Rotating about a horizontal axis → slices are vertical → integrate $dx$.
• Rotating about a vertical axis → slices are horizontal → integrate $dy$, which means rewriting the curves as $x$ in terms of $y$.

A region whose top boundary changes partway along needs two integrals in $x$ but often only one in $y$. Spending thirty seconds choosing can halve the work, and the choice is where the marks quietly are.