6.2 Determining Volumes by Slicing

Volumes of Solids

Volumes by cross-sectional integration

The slicing method finds the volume of a solid by integrating the area of its cross-sections along an axis. Think of it like slicing a loaf of bread: each slice has a measurable area, and adding up all those infinitely thin slices gives you the total volume.

Here's how to set it up:

1. Choose an axis to slice along (x-axis or y-axis).
2. Identify the cross-section shape at a generic point on that axis. This could be a rectangle, circle, triangle, semicircle, or any other shape.
3. Write the cross-section area as a function of your chosen variable.
4. Integrate that area function over the interval defined by the solid's boundaries.

The volume formula:

$V = \int_{a}^{b} A(x)\, dx \quad \text{or} \quad V = \int_{c}^{d} A(y)\, dy$

where $A(x)$ or $A(y)$ is the cross-sectional area at a given point, and the limits $a, b$ (or $c, d$) come from the solid's boundaries along the chosen axis.

A few examples to picture this:

• A rectangular prism sliced along the x-axis has constant rectangular cross-sections, so $A(x)$ is just a constant and the integral simplifies to area times length.
• A cone sliced perpendicular to its axis has circular cross-sections whose radii shrink linearly from base to tip. If the radius at position $x$ is $r(x)$, then $A(x) = \pi [r(x)]^2$.

The cross-sectional area perpendicular to the axis of integration is the core building block of every volume-by-slicing problem. Get that expression right, and the rest is just evaluating an integral.

Disk method for revolution solids

The disk method handles solids formed by revolving a region around an axis. Each cross-section perpendicular to the axis of revolution is a full disk (a filled-in circle), so you only need one function to define the radius.

Steps to set up a disk method integral:

1. Sketch the region bounded by the function(s) and identify the axis of revolution.
2. Find the disk radius at a generic point along the axis. The radius is the distance from the axis of revolution to the curve.
3. Write the cross-sectional area as $A = \pi r^2$, where $r$ is expressed in terms of your integration variable.
4. Integrate over the interval that spans the region.

Volume formulas:

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

Here $f(x)$ or $g(y)$ gives the radius of each disk, and the integration limits mark where the region starts and ends.

Example: Revolve the region under $y = x^2$ from $x = 0$ to $x = 1$ around the x-axis. At each $x$, the disk has radius $f(x) = x^2$, so:

$V = \pi \int_{0}^{1} (x^2)^2\, dx = \pi \int_{0}^{1} x^4\, dx = \pi \left[\frac{x^5}{5}\right]_0^1 = \frac{\pi}{5}$

A common mistake: make sure the radius is the distance from the curve to the axis of revolution, not just the function value. If you're revolving around $y = 3$ instead of $y = 0$, the radius changes to $3 - f(x)$.

Washer method for hollow solids

When the region being revolved doesn't touch the axis of revolution, or when it's bounded by two curves, each cross-section is a washer (a disk with a hole in the middle). You subtract the inner disk's area from the outer disk's area.

Steps to set up a washer method integral:

1. Sketch the region between the two curves and identify the axis of revolution.

2. Determine the outer radius $R$ (distance from the axis to the farther curve) and the inner radius $r$ (distance from the axis to the closer curve) at a generic point.

3. Write the cross-sectional area: $A = \pi(R^2 - r^2)$.

4. Integrate over the interval defined by the region's boundaries.

Volume formulas:

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

where $f$ is the outer function (farther from the axis) and $g$ is the inner function (closer to the axis).

Example: Revolve the region between $y = x$ and $y = x^2$ from $x = 0$ to $x = 1$ around the x-axis. Since $x \geq x^2$ on this interval, the outer radius is $R = x$ and the inner radius is $r = x^2$:

$V = \pi \int_{0}^{1} \left(x^2 - x^4\right) dx = \pi \left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1 = \pi\left(\frac{1}{3} - \frac{1}{5}\right) = \frac{2\pi}{15}$

Additional Volume Calculation Methods

The cylindrical shells method is an alternative to the disk and washer methods. Instead of slicing perpendicular to the axis of revolution, you integrate thin cylindrical shells that wrap around the axis. This method is especially useful when setting up a disk/washer integral would force you to solve for the other variable or split into multiple integrals.

Shells are typically covered in their own section, but the key idea to keep in mind: all of these volume methods rely on the same principle. You express an infinitesimal piece of volume (whether it's a thin disk, a washer, or a cylindrical shell), then use a definite integral to sum those pieces across the entire solid.

When choosing a method, consider:

• Disk method when the region touches the axis of revolution and is bounded by one curve.
• Washer method when there's a gap between the region and the axis, or the region is between two curves.
• Shells when integrating with respect to the variable perpendicular to the axis of rotation is simpler.

The region you're revolving determines everything: the functions in your integrand, the limits of integration, and which method will be cleanest to use.