➗Calculus II Unit 2 Review

Calculating volumes of solids is one of the most practical applications of integration. The core idea: slice a 3D solid into thin pieces whose areas you can calculate, then integrate those areas to recover the total volume. This section covers cross-sectional integration as the general framework, then the disk and washer methods as specialized tools for solids of revolution.

Volumes Using Cross-Sectional Integration

Every volume-by-slicing problem rests on one principle: if you know the area of every cross-section of a solid, you can integrate those areas along the slicing axis to get the volume.

Think of it like a loaf of bread. Each slice has some area $A(x)$ , and each slice has a tiny thickness $dx$ . Stack all those thin slices together and you get the full volume.

General formula:

$V = \int_{a}^{b} A(x)\, dx$

You can also slice along the $y$ -axis or $z$ -axis depending on the solid's orientation, giving $V = \int_{c}^{d} A(y)\, dy$ or $V = \int_{e}^{f} A(z)\, dz$ .

How to set up a cross-sectional volume problem:

Choose a slicing axis based on how the solid is oriented and which direction gives you cross-sections you can describe.
Identify the cross-section shape at a generic point along that axis (circle, square, equilateral triangle, semicircle, etc.).
Write the area as a function of position. For example, if the cross-sections are squares with side length $s(x)$ , then $A(x) = [s(x)]^2$ .
Determine the limits of integration from where the solid starts to where it ends along the axis.
Integrate $A(x)$ over that interval.

This method works for irregular solids that aren't solids of revolution. A classic textbook example: a solid whose base is a region in the $xy$ -plane and whose cross-sections perpendicular to the $x$ -axis are known shapes (squares, triangles, etc.).

Volumes using cross-sectional integration, Determining Volumes by Slicing · Calculus

Disk Method for Revolution Solids

The disk method is a special case of cross-sectional integration where every cross-section is a circle. It applies when you revolve a region around an axis and there's no gap between the solid and the axis.

When you revolve a curve around, say, the $x$ -axis, each cross-section perpendicular to the $x$ -axis is a disk (a filled-in circle). The radius of that disk equals the distance from the curve to the axis, which is just $f(x)$ .

Since the area of a circle is $\pi r^2$ , you get:

Revolving around the $x$ -axis: $V = \pi \int_{a}^{b} [f(x)]^2\, dx$
Revolving around the $y$ -axis: $V = \pi \int_{c}^{d} [g(y)]^2\, dy$

Setting up a disk method problem:

Sketch the region and identify the axis of revolution.
Confirm there's no gap between the region and the axis (if there is, you need the washer method).
Write the radius function. The radius is the distance from the curve to the axis of revolution. If you're revolving around the $x$ -axis, the radius is $r(x) = f(x)$ . If the axis is $y = k$ instead of $y = 0$ , the radius becomes $r(x) = f(x) - k$ (or $k - f(x)$ , whichever is positive).
Square the radius, multiply by $\pi$ , and integrate over the appropriate interval.

Common mistake: Forgetting to adjust the radius when the axis of revolution isn't the $x$ - or $y$ -axis itself. If you revolve around $y = -1$ , the radius isn't just $f(x)$ ; it's $f(x) + 1$ .

Washer Method for Hollow Solids

The washer method extends the disk method to handle regions that don't touch the axis of revolution, or regions bounded by two curves. Each cross-section is a washer (a disk with a hole in the middle).

You subtract the area of the inner disk from the outer disk:

$A = \pi R^2 - \pi r^2 = \pi(R^2 - r^2)$

where $R$ is the outer radius (farther curve from the axis) and $r$ is the inner radius (closer curve to the axis).

Revolving around the $x$ -axis: $V = \pi \int_{a}^{b} \left([f(x)]^2 - [g(x)]^2\right) dx$
Revolving around the $y$ -axis: $V = \pi \int_{c}^{d} \left([f(y)]^2 - [g(y)]^2\right) dy$

Here $f$ is the curve farther from the axis (giving the outer radius) and $g$ is the curve closer to the axis (giving the inner radius).

Setting up a washer method problem:

Sketch the region and identify both bounding curves and the axis of revolution.
Determine which curve is farther from the axis (outer radius $R$ ) and which is closer (inner radius $r$ ).
Express both radii as functions of the integration variable. If the axis isn't $y = 0$ or $x = 0$ , adjust both radii accordingly.
Integrate $\pi(R^2 - r^2)$ over the interval.

Common mistake: Writing $\pi \int (f - g)^2\, dx$ instead of $\pi \int (f^2 - g^2)\, dx$ . These are not the same thing. You square each function separately, then subtract.

Cylindrical Shell Method

The shell method is an alternative to disk/washer that's especially useful when setting up a disk or washer integral would force you to solve for $x$ in terms of $y$ (or vice versa) in an awkward way.

Instead of slicing perpendicular to the axis of revolution, you slice parallel to it. Each thin slice, when revolved, traces out a cylindrical shell rather than a disk.

The surface area of a thin shell with radius $r$ , height $h$ , and thickness $dx$ is $2\pi r \cdot h \cdot dx$ , so:

Rotation around the $y$ -axis: $V = 2\pi \int_{a}^{b} x\, f(x)\, dx$
Rotation around the $x$ -axis: $V = 2\pi \int_{c}^{d} y\, g(y)\, dy$

In each formula, the first factor after $2\pi$ is the shell radius (distance from the slice to the axis) and the second factor is the shell height (length of the slice).

When to use shells vs. disks/washers:

If the region is defined as $y = f(x)$ and you're revolving around the $y$ -axis, shells let you integrate with respect to $x$ directly, avoiding the need to invert the function.
If the region is defined as $y = f(x)$ and you're revolving around the $x$ -axis, the disk/washer method is usually more straightforward.
A good rule of thumb: if the axis of revolution is parallel to the variable you'd naturally integrate with respect to, shells tend to be easier. If it's perpendicular, disks/washers tend to be easier.

➗Calculus II Unit 2 Review