Problem Statement

Find the volume of the solid formed by rotating the region bounded by y = x², x = 0, and x = 2 around the y-axis.

Step 1: Identify the Setup

We are rotating around the y-axis, and our region is described in terms of x. This makes the shell method a natural fit — we'll integrate with respect to x.

The shell method formula for rotation around the y-axis is:

V = 2π ∫[a to b] x · f(x) dx

Step 2: Determine the Components

  • Radius of each shell: x (the horizontal distance from the y-axis)
  • Height of each shell: f(x) = x²
  • Limits of integration: x = 0 to x = 2

Step 3: Write the Integral

Substituting into the formula:

V = 2π ∫[0 to 2] x · x² dx

V = 2π ∫[0 to 2] x³ dx

Step 4: Evaluate the Integral

Integrate x³ with respect to x:

∫ x³ dx = x⁴/4

Apply the limits from 0 to 2:

[x⁴/4] from 0 to 2 = (2⁴/4) − (0⁴/4) = (16/4) − 0 = 4

Step 5: Multiply by 2π

V = 2π × 4 = 8π

The volume of the solid of revolution is 8π cubic units, which is approximately 25.13 cubic units.

Verification: Why Not the Disk Method?

We could also solve this using the disk method, but it requires integrating with respect to y. Since y = x², we'd need x = √y, and the limits would change to y = 0 to y = 4:

V = π ∫[0 to 4] (√y)² dy = π ∫[0 to 4] y dy = π [y²/2] from 0 to 4 = π(8) = 8π ✓

Both methods agree — but notice how the shell method kept everything in terms of x and avoided solving for the inverse function.

Key Takeaways

  • Always write radius and height explicitly before setting up the integral
  • The integrand for a shell around the y-axis is always x · f(x)
  • Multiply the evaluated integral by 2π at the end
  • Cross-checking with the disk method is a great way to verify your answer

Common Mistakes to Avoid

  • Forgetting the 2π factor — it's part of the circumference of the shell
  • Using the wrong limits — limits should match the variable of integration
  • Mixing up radius and height — radius is always the distance to the axis, height is the function value