Volumes: The Disk and Washer Method | Free Calculus 2 Lesson (4 College Credits) | UPI Study

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Volumes: The Disk and Washer Method

Calculus 2 · Chapter 4: Applications of IntegrationSection 4.2 · lesson 17 of 5410 min · free

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In this lesson

Building solids one thin slice at a time.

  • Use the cross-section principle V=\int A(x)\,dx
  • Apply the disk method when the region touches the axis
  • Apply the washer method when a gap creates a hole
  • Adapt the radii when revolving about any horizontal or vertical line

Transcript

Everything said in the video, section by section (1,541 words). Read along, or search it.

MAT-265 - Calculus 2 · Section 4.2

Volumes: Disk and Washer

A wooden bowl on a lathe. A turned chair leg. A wine glass. A car's brake rotor. Each one is a solid of revolution — formed by spinning a flat shape around an axis. A manufacturer who needs the weight of a turned metal part, or a designer estimating how much liquid a curved vessel holds, has to compute the volume of exactly such a solid. And there's no single formula for these shapes, the way there is for a box or a sphere. Because the profile can be any curve at all. So integration solves the problem by the oldest trick in calculus. Cut the solid into a great many thin slices whose volumes are easy. Then add them up. Section four point one found areas by summing thin rectangles. This section extends exactly the same idea into three dimensions.

The Parent Idea

Volume by Cross-Section

The foundation for this section, and the next one, is a single principle. Imagine slicing a solid into thin slabs perpendicular to some axis. Like cutting a loaf of bread. If the slice taken at position x has cross-sectional area A of x, and thickness d x, then its volume is approximately A of x, times d x. Sum the slices, pass to the limit, and the sum becomes an integral. So the volume is the integral from a to b of A of x, d x. And here's what makes this powerful. The whole task of any volume problem reduces to one question. What is the cross-sectional area? For solids of revolution, that cross-section is always a circle, or a ring. Which makes A of x very easy to write down.

No Gap

The Disk Method

Start with the simpler case. When a region lies flush against the axis of revolution — with no gap between them. Revolving it produces a solid whose cross-sections are full circles. Disks. A representative thin strip of the region, of height f of x and width d x, sweeps out a thin disk of radius r equals f of x. The disk's circular face has area pi r squared, which is pi times f of x, squared. And its thickness is d x. So its volume is pi f of x squared, d x. That's just the cross-section principle, with A of x equal to pi f of x squared. So the disk formula is: V equals pi, times the integral from a to b, of f of x squared, d x. Let's use it. The region under y equals root x, from zero to four, revolved about the x-axis. The region sits directly on the x-axis, so the disk method applies. And f of x squared is just x. So V is pi times the integral of x, from zero to four. Which is pi times sixteen over two. Eight pi. About twenty-five point one three cubic units. Now — that squaring step is where the pi and the radius come together. And forgetting to square f of x, or forgetting the factor pi entirely, are the two most frequent slips in disk problems. Watch for both.

With a Gap

The Washer Method

Now the case with a gap. When the region being revolved does not touch the axis — there's a gap, or the region lies between two curves — the resulting cross-section is not a full disk. It's a ring. Called a washer. Each washer is a large disk of outer radius R, with a smaller disk of inner radius r removed from its center. So its area is the difference of the two circular areas. Pi R squared, minus pi r squared. Which is pi times R squared minus r squared. Summing those ring areas gives the washer formula. Now here is the single most important warning in this section. The squares are subtracted. Not the radii. It is pi times the integral of R squared minus r squared. Never pi times the integral of R minus r, squared. Squaring the difference would be a serious error — and geometrically it describes something else entirely. Let's do one. The region bounded by y equals x and y equals x squared, revolved about the x-axis. On zero to one, the line lies above the parabola. So the outer radius is x, and the inner radius is x squared. R squared is x squared. And r squared is x squared, squared — which is x to the fourth. Each squared separately, then subtracted. The integral gives pi times one third minus one fifth. Which is two pi over fifteen. About zero point four one nine.

Figures 4.2a and 4.2b

What the Slice Actually Sweeps Out

Let's look at what a slice actually sweeps out, because the picture settles which method you need. On the left, the disk case. The region is flush against the axis. What you're seeing is the silhouette of the solid — the region, plus its mirror image below the axis. The red bar is the representative strip. Revolve it, and it sweeps out a solid disk of radius f of x. No hole anywhere. On the right, the washer case. Here a second curve separates the region from the axis. So when you revolve, that gap becomes a hole running through the middle. You can see the white lens in the center — that's the hole. Material that isn't there. The slice is now a ring, with outer radius R and inner radius r. And the hole's area gets subtracted from the outer disk's area. Look at the two pictures and ask which one your region matches. That's the whole decision.

Choosing the Axis

Radius Always Means Distance From the Axis

So far we've revolved about the x-axis. But both methods adapt to any horizontal or vertical axis. Only the radius expressions change. Because radius always means distance from the axis. Nothing else. Three situations cover nearly every problem. First. About the y-axis. Slice horizontally instead of vertically. Solve the boundary curves for x as functions of y, integrate in y, and measure radii horizontally from the y-axis. Second. About a horizontal line y equals k. Here the radius is the vertical distance from that line to the curve. Absolute f of x minus k. Not f of x itself. That's the step students skip, and it's why their answer is wrong. Third. About a vertical line x equals h. By symmetry with the previous case, the radius is the horizontal distance, absolute g of y minus h. And you integrate in y. One rule ties it together. Match the slice to the axis. Vertical strips, integrating in x, for horizontal axes. Horizontal strips, integrating in y, for vertical axes. A slice has to be perpendicular to the axis it revolves around.

Worked Example 3

Shifting the Axis Changes the Method

Let's see how much the axis matters. Take the same region as before. Under y equals root x, from zero to four. But now revolve it about the line y equals minus one, instead of the x-axis. Revolving about y equals minus one creates a gap of width one, between the region and the axis. So a hole appears. And the washer method is now needed — for a region that a moment ago needed only disks. The outer radius reaches from y equals minus one, up to the curve. So R is root x, minus minus one. Which is root x plus one. The inner radius reaches from y equals minus one, up to the region's lower edge at y equals zero. So r is one. Set up the washer integral, expand, and you get pi times the integral of x plus two root x. Which evaluates to fifty-six pi over three. About fifty-eight point six four cubic units. So shifting the axis away from the region turned a disk problem into a washer problem. A reminder that the choice of axis — not just the curve — determines which method applies. One thing to think about, and it's the deepest point in this section. In Example two the radii were R equals x and r equals x squared, and the volume used R squared minus r squared. Suppose a classmate instead computes pi times the integral of x minus x squared, all squared — reasoning that the strip's length is x minus x squared. Explain geometrically why that gives the wrong volume. And ask yourself: what solid would pi times the integral of R minus r, squared, actually describe? If any.

Key Takeaways

What to Carry Forward

Five things to carry forward. Every volume in this section comes from one idea. V equals the integral of A of x, where A of x is the cross-sectional area. Use the disk method when the region touches the axis, so there's no hole. V equals pi times the integral of f of x squared. Use the washer method when a gap separates the region from the axis, so a hole appears. V equals pi times the integral of R squared minus r squared. Subtracting the squares. Never squaring the difference. Radius always means distance from the axis. Revolving about y equals k, or x equals h, shifts every radius accordingly. And match the slice to the axis. Vertical strips for horizontal axes. Horizontal strips for vertical axes. Next up is section four point three — the Cylindrical Shell Method, which slices parallel to the axis instead of perpendicular to it. And that turns out to solve problems the washer method finds awkward.

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FAQ

Questions students ask us

About Volumes: The Disk and Washer Method

What is the disk method in calculus?
The disk method finds the volume of a solid of revolution when the region lies flush against the axis, so every cross-section is a full circle. A strip of height f(x) sweeps out a disk of area \pi[f(x)]^2, which gives V=\pi\int_a^b [f(x)]^2\,dx. For example, revolving y=\sqrt{x} on [0,4] about the x-axis gives V=\pi\int_0^4 x\,dx=8\pi\approx 25.13 cubic units.
When do you use the washer method instead of the disk method?
Use the washer method when a gap separates the region from the axis of revolution, so the solid has a hole and each cross-section is a ring. The volume is V=\pi\int_a^b (R^2-r^2)\,dx, where R is the outer radius and r is the inner radius. For example, revolving the region between y=x and y=x^2 about the x-axis gives \pi\int_0^1 (x^2-x^4)\,dx=\frac{2\pi}{15}\approx 0.419.
How do you find the radius when revolving around a line like y = -1?
The radius is always the distance from the axis, so for an axis y=k use |f(x)-k| rather than f(x) alone. Revolving the region under y=\sqrt{x} on [0,4] about y=-1 creates a gap, so R=\sqrt{x}+1 and r=1. Then V=\pi\int_0^4 (x+2\sqrt{x})\,dx=\frac{56\pi}{3}\approx 58.64. For a vertical axis x=h, use horizontal slices with radius |g(y)-h| and integrate in y.
What is the most common mistake with the washer method?
The most common mistake is squaring the difference of the radii instead of subtracting their squares. The correct integrand is \pi(R^2-r^2), never \pi(R-r)^2, because a washer's area is the outer disk's area minus the area of the hole. Other frequent slips include forgetting to square f(x), dropping the factor \pi, and using f(x) as the radius when the axis is not the x-axis.

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