How to Find Volume by Spinning: Disk Method

This method leverages the symmetry of revolving shapes, transforming a 2D area into a 3D object. Each disk’s volume is \( \pi r^2 \Delta x \), where \( r \) is the radius function. By accumulating these volumes along the axis, complex volumetric problems are simplified. Especially useful in engineering and design, the Disk Method provides a foundational tool for calculating storage capacities, structural volumes, and more.

Here are the steps to use the method:

1. Define the Radius: The radius \(r\) of each disk is a function of \(x\) (or \(y\), depending on the axis of revolution), denoted as \(r(x)\) or \(r(y)\).
2. Disk Volume: The volume of each infinitesimally thin disk is given by the formula \(dV = \pi [r(x)]^2 dx\) (if revolving around the y-axis) or \(dV = \pi [r(y)]^2 dy\) (if revolving around the x-axis), where \(dV\) is the differential volume of the disk.
3. Set Up the Integral: The total volume \(V\) of the solid is the integral of these differential volumes over the interval of revolution. Formally, if revolving around the y-axis:
   \( V = \int_{a}^{b} \pi [r(x)]^2 dx \)
   And if revolving around the x-axis:
   \( V = \int_{c}^{d} \pi [r(y)]^2 dy \)
   Here, \(a\) and \(b\) (or \(c\) and \(d\)) are the bounds of the region being revolved, determined by the intersection points of the curve and the axis of revolution or the specified limits of integration.
4. Evaluate the Integral: Compute the integral to find the total volume of the solid of revolution. This step may require analytical techniques or numerical methods depending on the function defining the radius.

By applying this method, you can calculate the volume of a wide variety of shapes with rotational symmetry, providing a powerful technique for analyzing physical objects and theoretical constructs in mathematics, physics, and engineering.

Let’s calculate the volume of a solid formed by revolving the region under the curve \( y = \sin(x) \) from \( x = 0 \) to \( x = \pi \) around the x-axis.

Step-by-Step Solution:

1. Define the Radius:

• The radius of each disk is given by the function \( y = \sin(x) \).

2. Disk Volume:

• The volume of each disk is \( dV = \pi [r(x)]^2 dx = \pi [\sin(x)]^2 dx \).

3. Set Up the Integral:

• The total volume \( V \) is the integral of \( \pi [\sin(x)]^2 \) from \( x = 0 \) to \( x = \pi \):
  \( V = \int_{0}^{\pi} \pi [\sin(x)]^2 dx \)

4. Simplify the Integral Using a Trigonometric Identity:

• Use the identity \( \sin^2(x) = \frac{1 – \cos(2x)}{2} \) to simplify the integral:
  \( V = \pi \int_{0}^{\pi} \frac{1 – \cos(2x)}{2} dx \)

The volume of the solid formed by revolving the region under the curve \( y = \sin(x) \) from \( x = 0 \) to \( x = \pi \) around the x-axis is approximately \(4.93\) cubic units.

Recommended EffortlessMath Books

For the calculus background behind volumes of revolution, the Pre-Calculus for Beginners pins down the function and integration prerequisites, and the Calculus for Beginners walks through every integration technique – including disk, washer, and shell methods – with full worked examples.

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Frequently Asked Questions

What is the disk method?

A way to find the volume of a solid of revolution by adding up thin disks (flat circles). When you rotate a region around an axis, each vertical strip becomes a disk. Sum all the disks with an integral: \(V=\pi\int_a^b [f(x)]^2\,dx\) for rotation about the \(x\)-axis.

When should I use the disk method?

Use disks when the region is up against the axis of rotation – no gap between the region and the axis means no hole in the solid. If there IS a gap, use the washer method (which subtracts an inner disk from an outer one).

What’s the formula for rotation about the x-axis?

\(V=\pi\int_a^b [f(x)]^2\,dx\), assuming \(f(x)\geq 0\) on \([a,b]\). The radius of each disk is \(f(x)\), so its area is \(\pi [f(x)]^2\). Multiply by thickness \(dx\) and integrate. The \(\pi\) is constant and pulls outside the integral. Example: rotating \(y=x\) from 0 to 3 about the \(x\)-axis gives \(V=\pi\int_0^3 x^2\,dx=9\pi\).

What about rotation about the y-axis?

Rewrite the function as \(x=g(y)\) and integrate in \(y\): \(V=\pi\int_c^d [g(y)]^2\,dy\), where \(c\) and \(d\) are the bounds in \(y\). The disks now sit horizontally, perpendicular to the \(y\)-axis. Example: rotating \(y=\sqrt{x}\) (so \(x=y^2\)) from \(y=0\) to \(y=2\) about the \(y\)-axis gives \(V=\pi\int_0^2 y^4\,dy=\frac{32\pi}{5}\).

Walk through a worked example?

Rotate \(y = \sqrt{x}\) from \(x=0\) to \(x=4\) about the \(x\)-axis. Each disk has radius \(\sqrt{x}\), so \(r^2 = x\). \(V=\pi\int_0^4 x\,dx = \pi[\frac{x^2}{2}]_0^4 = \pi\cdot 8 = 8\pi\). Volume is \(8\pi\) cubic units.

What’s the difference between disk and washer methods?

Disk: solid disks, no hole. Washer: disks with a hole (think washer or donut). Use washers when the region is NOT touching the axis. Washer formula: \(V=\pi\int_a^b [(R(x))^2 – (r(x))^2]\,dx\), where \(R\) is the outer radius and \(r\) is the inner.

What about rotating around a horizontal line that’s not the x-axis?

The radius of each disk is the distance from the function to the line. Rotating \(y=f(x)\) about \(y=k\): radius \(=|f(x)-k|\). Volume \(=\pi\int_a^b (f(x)-k)^2\,dx\). Same idea, shifted axis. Example: rotating \(y=x^2\) about \(y=-1\) for \(0 \leq x \leq 1\) gives radius \(x^2 + 1\), so \(V=\pi\int_0^1 (x^2+1)^2\,dx\).

How do I find the bounds of integration?

The bounds are the \(x\)-values (or \(y\)-values) where the region starts and ends. If the region is bounded by curves, find where they intersect to get the bounds. For \(y=x^2\) and \(y=4\), they intersect at \(x=\pm 2\), so bounds are \(-2\) and \(2\) when rotating about the \(x\)-axis.

Can I just remember one formula and adapt?

Yes – remember \(V=\pi\int (\text{radius})^2\,d(\text{axis variable})\). Plug in the right radius for your setup. Rotating about \(x\)-axis: \(d(\text{axis variable}) = dx\), radius is \(f(x)\). Rotating about \(y\)-axis: \(dy\), radius is \(g(y)\).

Where does the disk method show up on tests?

AP Calculus AB and BC, college Calc I and II. Free-response questions often give a region and ask for volume of revolution – your job is to choose disk, washer, or shell based on the setup. Disk is the simplest; learn it first, then washer (with the hole), then shell.

Related EffortlessMath Lessons

If a topic on this page feels rusty, these short lessons go deeper:

• Volume by spinning: shell method
• Volume by spinning: washer method
• Volume by cross-sections
• How to calculate the volume of cubes and prisms
• How to find the area of a quarter circle