Source: PS3 worksheet
Tags: volume of revolution, shell method, disk method, washer method, solids of revolution, y = 1/(x² + 4), rotation around y-axis, rotation around x-axis, Calc 2
Difficulty: Intermediate | Prerequisites: Calc I integration, understanding of area between curves, basic trigonometric substitution
Volumes of revolution are one of the major applications of integration in Calc II. The idea is simple: take a 2D region, spin it around an axis, and compute the volume of the resulting 3D solid. There are two main methods (shell and disk/washer), and choosing the right one depends on the axis of rotation and which variable is more convenient to integrate over. These two problems use the same region but different axes, so they let you practise both methods side by side.
The shell method integrates cylindrical shells (2π × radius × height × thickness). The disk/washer method integrates circular cross-sections (π × radius²). For y = 1/(x² + 4) between x = 0 and x = 5, rotating around the y-axis favours shells (integrate in x), while rotating around the x-axis favours disks (also integrate in x). Each problem asks you to do it both ways.
Volume of revolution (solid of revolution)
The 3D solid formed by rotating a 2D region around a line (the axis of revolution). Think of it as spinning a flat cutout on a potter's wheel.
Shell method (cylindrical shells)
An integration technique where you slice the solid into thin cylindrical shells parallel to the axis of rotation. The volume element is 2π × (shell radius) × (shell height) × dx (or dy).
Disk method
An integration technique where you slice the solid into thin circular disks perpendicular to the axis of rotation. The volume element is π × (radius)² × dx (or dy). Used when the region has no hole in the middle.
Washer method
A variation of the disk method for regions with a hole: each slice is a ring (washer) with an outer radius and an inner radius. The volume element is π × (R_outer² - R_inner²) × dx (or dy).
Axis of revolution
The line around which the region is rotated. Changing the axis (x-axis vs y-axis, or an offset line) changes which method is simpler and how you set up the integral.
Setup
The region is bounded by y = 1/(x² + 4), y = 0 (the x-axis), x = 0 (the y-axis), and x = 5. Rotating this around the y-axis.
With shells, each shell is a vertical strip at position x, with:
Radius = x (distance from the y-axis)
Height = 1/(x² + 4) (the function value)
Thickness = dx
x ranges from 0 to 5
Integral
V = \int_0^5 2\pi x \cdot \frac{1}{x^2 + 4}\, dx = 2\pi \int_0^5 \frac{x}{x^2 + 4}\, dxEvaluation
Use the substitution u = x² + 4, so du = 2x dx, meaning x dx = du/2.
When x = 0: u = 4
When x = 5: u = 29
V = 2\pi \int_4^{29} \frac{1}{u} \cdot \frac{du}{2} = \pi \int_4^{29} \frac{du}{u} = \pi \left[\ln u\right]_4^{29} = \pi(\ln 29 - \ln 4) = \pi \ln\frac{29}{4}Result
V = π ln(29/4), which is approximately 6.27 cubic units.
Setup
Rotating the same region around the y-axis, but now slicing horizontally (in dy). Each horizontal slice is a washer.
First, invert y = 1/(x² + 4) to express x in terms of y:
x² + 4 = 1/y, so x² = 1/y - 4, giving x = √(1/y - 4).
Key y-values
At x = 0: y = 1/4 (the top of the curve on the y-axis)
At x = 5: y = 1/29 (the right endpoint)
The region needs to be split into two parts based on whether the outer boundary is x = 5 or x = √(1/y - 4).
Two zones
For y from 0 to 1/29: the horizontal line at height y does not intersect the curve (the curve stays above y = 1/29 for 0 <= x <= 5). The washer has outer radius 5 and inner radius 0 (a full disk).
For y from 1/29 to 1/4: the horizontal line at height y intersects the curve at x = √(1/y - 4). The washer has outer radius √(1/y - 4) and inner radius 0.
Integral
V = \int_0^{1/29} \pi(5)^2\, dy + \int_{1/29}^{1/4} \pi\left(\frac{1}{y} - 4\right) dyEvaluation
First integral: π(25)(1/29) = 25π/29.
Second integral: π [ln(1/4) - 4(1/4) - ln(1/29) + 4(1/29)] = π [ln(29/4) - 4(1/4 - 1/29)] = π [ln(29/4) - 100/116] = π [ln(29/4) - 25/29].
Result
V = 25π/29 + π[ln(29/4) - 25/29] = π ln(29/4).
This matches the shell method answer, which is a good check.
Setup
Same region, now rotated around the x-axis. Slicing vertically (in dx), each slice is a disk with:
Radius = y = 1/(x² + 4)
Thickness = dx
x ranges from 0 to 5
Integral
V = \int_0^5 \pi \left(\frac{1}{x^2 + 4}\right)^2 dx = \pi \int_0^5 \frac{1}{(x^2 + 4)^2}\, dxEvaluation via trigonometric substitution
Let x = 2 tan θ, so dx = 2 sec² θ dθ and x² + 4 = 4 sec² θ.
When x = 0: θ = 0
When x = 5: θ = arctan(5/2)
Substituting:
V = \pi \int_0^{\arctan(5/2)} \frac{2\sec^2\theta}{16\sec^4\theta}\, d\theta = \frac{\pi}{8} \int_0^{\arctan(5/2)} \cos^2\theta\, d\thetaUsing the identity cos²θ = (1 + cos 2θ)/2:
V = \frac{\pi}{16} \left[\theta + \frac{\sin 2\theta}{2}\right]_0^{\arctan(5/2)}For θ = arctan(5/2): from a right triangle with opposite = 5, adjacent = 2, hypotenuse = √29, we get sin θ = 5/√29, cos θ = 2/√29, so sin 2θ = 2(5/√29)(2/√29) = 20/29.
Result
V = \frac{\pi}{16}\left(\arctan\frac{5}{2} + \frac{10}{29}\right)Setup
Rotating around the x-axis, using shells means slicing horizontally (in dy). Each shell is a horizontal strip at height y, with:
Radius = y (distance from the x-axis)
Height = the horizontal width of the region at that y-level
Thickness = dy
Determining shell height
The curve is y = 1/(x² + 4), so solving for x: x = √(1/y - 4), valid for y <= 1/4.
The horizontal extent depends on the y-level (same two-zone logic as the washer method for the y-axis):
For 0 < y <= 1/29: the curve does not limit the region at this height. The strip runs from x = 0 to x = 5, so the height (width) = 5.
For 1/29 < y <= 1/4: the strip runs from x = 0 to x = √(1/y - 4), so the height (width) = √(1/y - 4).
Integral (setup, not evaluated)
V = \int_0^{1/29} 2\pi y \cdot 5\, dy + \int_{1/29}^{1/4} 2\pi y \sqrt{\frac{1}{y} - 4}\, dyThe problem only asks for the setup, not evaluation. This integral is considerably harder to compute than the disk method version, which is precisely the point: some axis/method combinations are much easier than others.
Shell method (general)
V = \int_a^b 2\pi (\text{radius})(\text{height})\, d(\text{variable})Disk method (general)
V = \int_a^b \pi [R(x)]^2\, dxWasher method (general)
V = \int_a^b \pi \left([R_{\text{outer}}]^2 - [R_{\text{inner}}]^2\right) d(\text{variable})Trig substitution identity used here
For integrals involving (x² + a²), let x = a tan θ. Also: cos²θ = (1 + cos 2θ)/2.
Volumes of revolution are how engineers calculate the volume of axially symmetric parts: pipes, nozzles, turbine blades, wine glasses, vases, and any turned object. The same integrals appear in calculating moments of inertia for rotating machinery.
Students often confuse which method to use. A reliable rule: if your shells are parallel to the axis of rotation, use the shell method. If your slices are perpendicular to the axis, use the disk/washer method.
Forgetting the 2π in the shell formula is a frequent error. The shell formula always has a factor of 2π (the circumference of the shell).
When inverting y = 1/(x² + 4) to set up the washer integral in y, students forget to split the region into two zones. If you do not separate the portion below y = 1/29 from the portion above, you get the wrong limits or the wrong radii.
Mixing up radius and height in the shell method. The radius is the distance from the shell to the axis of rotation, not the function value.
⚠️ Exams almost always include at least one volume-of-revolution problem. Being fluent in both methods is expected.
⚠️ The trig substitution for 1/(x² + a²)² is a classic Calc II technique. Practise it until it is automatic.
⚠️ When both methods give the same answer, you know you have set them up correctly. Verifying with the alternate method is strong exam strategy.
True or false: The shell method slices perpendicular to the axis of rotation.
False. Shells are parallel to the axis.
Fill in the blank: The shell method volume element is 2π × ______ × height × thickness.
Radius (distance from the axis of rotation).
True or false: When rotating around the y-axis and integrating in x, the shell method is typically easier than the washer method.
True. The shell method avoids inverting the function.
Fill in the blank: For the substitution x = 2 tan θ, we get x² + 4 = ______.
4 sec²θ.
Q: Set up (but do not evaluate) the shell method integral for the volume obtained by rotating y = 1/(x² + 4), y = 0, x = 0, x = 5 around the y-axis.
A: V = 2π ∫₀⁵ x/(x² + 4) dx.
Q: Evaluate the integral 2π ∫₀⁵ x/(x² + 4) dx using the substitution u = x² + 4.
A: u = x² + 4, du = 2x dx. Limits: 4 to 29. The integral becomes π ∫₄²⁹ (1/u) du = π[ln u]₄²⁹ = π ln(29/4).
Q: When rotating the same region around the x-axis, which method gives a simpler integral, shell or disk?
A: The disk method. The integral π ∫₀⁵ 1/(x² + 4)² dx requires a trig substitution but stays in one piece. The shell method requires splitting into two zones and produces a harder integrand.
Q: What substitution do you use for ∫ 1/(x² + 4)² dx?
A: x = 2 tan θ. This converts (x² + 4)² into (4 sec²θ)² = 16 sec⁴θ, and dx = 2 sec²θ dθ.
The u-substitution used in the shell integral connects back to Calc I integration techniques. The trig substitution for the disk integral previews material from later in Calc II (integration techniques chapter). Both methods also connect to surface area of revolution, which uses similar setups but with arc length elements instead of height.
volume of revolution, solid of revolution, shell method, disk method, washer method, cylindrical shells, rotation around y-axis, rotation around x-axis, trig substitution, u-substitution, 1/(x² + 4), arctan substitution, Calc 2 NYU, Calculus II for Engineers, volumes by integration