Volumes: The Cylindrical Shell Method | Free Calculus 2 Lesson (4 College Credits) | UPI Study

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Volumes: The Cylindrical Shell Method

Calculus 2 · Chapter 4: Applications of IntegrationSection 4.3 · lesson 18 of 5410 min · free

← Volumes: The Disk and Washer MethodNext lesson: Arc Length of a Curve →

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In this lesson

Slicing parallel to the axis instead of across it.

  • Derive dV = 2\pi r h\,dx by unrolling a shell
  • Recognise when shells beat washers
  • Adapt the radius for any parallel axis
  • Switch to dy for a horizontal axis

Transcript

Everything said in the video, section by section (1,496 words). Read along, or search it.

MAT-265 - Calculus 2 · Section 4.3

The Cylindrical Shell Method

Picture the region trapped between the curve y equals two x minus x squared, and the x-axis. Now imagine spinning it around the y-axis, to sweep out a solid. Something like a smooth, rounded bundt pan. Try to find its volume with the washer method, and you hit a wall immediately. Washers slice horizontally. Which means you'd have to solve y equals two x minus x squared, for x in terms of y. That equation is a quadratic in x. Solving it drags in a square root — and it forces the region to split into two pieces. The calculation becomes a tangle before any integration even begins. So there's a better way. And it comes from changing which direction you slice.

The Problem

A Stubborn Volume

Instead of slicing the solid into flat washers stacked along the axis, do this. Slice the region into thin vertical strips — parallel to the axis of revolution. Then spin each strip into a thin hollow tube. A cylindrical shell. Like the concentric layers of an onion. Or a set of nested tin cans. Summing the volumes of all those nested shells gives the total volume. And crucially — it keeps the function in its original, convenient form. y equals two x minus x squared. No rearranging. Here's the single structural difference, and it's worth holding onto for the rest of the chapter. Washers slice perpendicular to the axis. Shells slice parallel to it. Everything else in this section follows from that one contrast.

The Derivation

Unroll the Shell

The heart of the method is the volume of one thin shell. So let's derive it. Consider a representative vertical strip of the region, at horizontal position x, with small width d x and height h. When this strip is revolved about a vertical axis, it sweeps out a cylindrical shell. A tube of radius r — the distance from the axis to the strip. Height h. And wall thickness d x. Now here's the key insight, and it's genuinely elegant. Imagine slitting that shell vertically, and unrolling it flat. You get a thin slab. And the slab's dimensions are easy to read off. Its length is the distance around the shell — the circumference, two pi r. Its height is the shell height, h. And its thickness is the strip width, d x. So the volume of the slab — and therefore of the thin shell — is just the product of those three. d V equals two pi r, times h, times d x. That approximation becomes exact in the limit as the thickness shrinks to zero.

The Formula

Sum the Nested Shells

Summing — that is, integrating — the volumes of all shells from x equals a to x equals b gives the shell formula. If a region is revolved about a vertical axis, and a representative strip at position x has radius r of x and height h of x, then the volume is the integral of two pi r h, d x. For revolution about the y-axis specifically, the radius is simply x, and the height is f of x. So the formula becomes the integral of two pi x f of x, d x. Let's go back and resolve that stubborn volume from the opening. Revolve the region under y equals two x minus x squared, from zero to two, about the y-axis. The strip at position x has radius x, and height two x minus x squared. Apply the formula. You get the integral of two pi x, times two x minus x squared. Distribute, and that's two pi times the integral of two x squared minus x cubed. Integrate term by term: two x cubed over three, minus x to the fourth over four. Evaluate at the bounds, and you get two pi times four thirds. Which is eight pi over three. And notice what we never did. We never solved the curve for x. The whole problem that made washers a tangle simply didn't arise.

Choosing a Method

When Shells Beat Washers

Both the shell and the washer method compute the same volumes. So neither one is universally correct. The practical question is which one leads to the simpler integral. And the deciding factor is the relationship between the slicing direction and the axis of revolution. From that single difference — washers perpendicular, shells parallel — flow two useful rules of thumb. First. Match the variable to the curve's natural form. If the region is bounded by functions of x — curves written y equals f of x — and the axis is vertical, then shells integrate in x and avoid solving for x entirely. Washers would force a rewrite into x equals g of y. Second. Watch for region-splitting. If horizontal slices would cut across different boundary curves at different heights, requiring the integral to be broken into pieces, shells often keep everything in one clean integral. And the reverse happens too. When the axis is vertical but the region is most simply described by functions of y, washers in y may win. Here's a reliable mental shortcut that captures all of it. A representative rectangle drawn parallel to the axis of revolution becomes a shell. A rectangle drawn perpendicular to the axis becomes a washer. So before choosing a method: sketch the region, draw the axis, and draw one representative rectangle in the direction that keeps your integration variable matched to how the curves are written. The rectangle's orientation tells you which method to use — and which variable to integrate in.

Axis Variations

Only the Radius Changes

The shell formula adapts to any axis parallel to the strips. And only one thing changes. The radius expression. Because the radius is always the distance from the axis to the strip. The height expression is completely unaffected by where the axis lies. Consider revolution about a vertical line x equals c, rather than the y-axis. The radius of the strip at position x becomes the horizontal distance, absolute x minus c. For a region lying to the right of the axis, that's x minus c. For a region to the left, it's c minus x. Let's do one. Revolve the region under y equals x squared, from zero to two, about the vertical line x equals three. Each strip sits to the left of the axis. So its radius is the distance three minus x. And the height is still x squared. Unchanged. Set up the integral: two pi, times three minus x, times x squared. Expand and integrate, and you get two pi times eight minus four. Which is eight pi. Notice how only the radius changed from the y-axis case — from x, to three minus x. The height stayed exactly where it was.

Horizontal Axes

Strips Turn Sideways, and So Does the Integral

One more variation. What if the axis is horizontal? If the axis is the x-axis, or a line y equals d, then the strips must be drawn horizontally — to stay parallel to the axis. So the integration switches to y. The radius becomes a vertical distance. And the height becomes a horizontal length, expressed in terms of y. But the structure of the formula is identical. Two pi, times radius, times length. Let's do one. The region bounded by y equals root x, y equals zero, and x equals four, revolved about the x-axis. Using shells. Slice horizontally with strips of thickness d y. The curve y equals root x rewrites as x equals y squared. And y ranges from zero to two, since x equals four gives y equals two. A horizontal strip at height y has radius y — that's the distance up from the x-axis. And its length is four minus y squared. That's the horizontal distance from the curve rightward to the line x equals four. Apply the formula and integrate. You get two pi times eight minus four. Eight pi again. Washers in x would also work for this one. The shell version is shown just to illustrate the horizontal-axis pattern.

Key Takeaways

What to Carry Forward

Five things to carry forward. A thin cylindrical shell has volume two pi r h d x. Derived by unrolling it into a slab of length two pi r, height h, and thickness d x. The shell method integrates those. V equals the integral of two pi r of x, h of x, d x — with strips drawn parallel to the axis of revolution. Shells are preferable when the region's curves are written to match the integration variable, and washers would force solving for the other variable or splitting the region. Most commonly, that means revolving a y equals f of x region about a vertical axis. For an axis x equals c, the radius is the distance absolute x minus c. And only the radius changes when the axis moves. Never the height. Revolving about a horizontal axis switches the strips, and the integration, to y. Radius becomes a vertical distance, height becomes a horizontal length in terms of y. Next up is section four point four — Arc Length. Where instead of measuring the region a curve bounds, we measure the curve itself.

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FAQ

Questions students ask us

About Volumes: The Cylindrical Shell Method

What is the cylindrical shell method for finding volume?
The cylindrical shell method finds the volume of a solid of revolution by slicing the region into thin strips parallel to the axis and adding up the hollow tubes they sweep out. Unrolling one shell gives a thin slab of length 2\pi r, height h, and thickness dx, so dV = 2\pi r h\,dx. Integrating gives V = \int_a^b 2\pi r(x)h(x)\,dx.
When should you use the shell method instead of the washer method?
Use shells when the curves are written as y = f(x) and the axis of revolution is vertical, because washers would force you to solve for x in terms of y or split the region. For example, revolving the region under y = 2x - x^2 from 0 to 2 about the y-axis gives \int_0^2 2\pi x(2x - x^2)\,dx = \frac{8\pi}{3} with no rearranging. A rectangle parallel to the axis becomes a shell, and a perpendicular one becomes a washer.
How do you find the radius in the shell method when rotating around a line like x = 3?
For a vertical axis x = c, the radius is the distance |x - c|. Use x - c if the region lies to the right of the axis and c - x if it lies to the left. Only the radius changes, and the height stays the same. Revolving the region under y = x^2 from 0 to 2 about x = 3 gives \int_0^2 2\pi(3 - x)x^2\,dx = 8\pi.
How do you use the shell method around a horizontal axis like the x-axis?
Draw the strips horizontally so they stay parallel to the axis, and integrate with respect to y. The radius becomes a vertical distance, and the height becomes a horizontal length written in terms of y. For the region bounded by y = \sqrt{x}, y = 0, and x = 4 revolved about the x-axis, rewrite the curve as x = y^2. Then V = \int_0^2 2\pi y(4 - y^2)\,dy = 8\pi.

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