Problem Statement
Find the volume of the solid formed by rotating the region bounded by y = 2x − x² and y = 0 around the y-axis.
Step 1: Find the Limits of Integration
First, determine where y = 2x − x² intersects y = 0:
2x − x² = 0
x(2 − x) = 0
So x = 0 and x = 2.
The region spans from x = 0 to x = 2.
Step 2: Confirm the Shell Method Setup
We're rotating around the y-axis, and our function is given in terms of x. The shell method is ideal here. The components are:
- Radius: x (distance from shell to y-axis)
- Height: f(x) − g(x) = (2x − x²) − 0 = 2x − x²
- Limits: 0 to 2
Step 3: Set Up the Integral
V = 2π ∫[0 to 2] x · (2x − x²) dx
Expand the integrand:
x · (2x − x²) = 2x² − x³
V = 2π ∫[0 to 2] (2x² − x³) dx
Step 4: Evaluate the Integral
Find the antiderivative term by term:
- ∫ 2x² dx = (2x³)/3
- ∫ x³ dx = x⁴/4
So: ∫(2x² − x³) dx = (2x³)/3 − x⁴/4
Apply the limits x = 0 to x = 2:
= [(2·8)/3 − 16/4] − [0 − 0]
= [16/3 − 4]
= [16/3 − 12/3]
= 4/3
Step 5: Final Volume
V = 2π · (4/3) = 8π/3
The volume is 8π/3 cubic units, approximately 8.38 cubic units.
Visual Interpretation
The curve y = 2x − x² is a downward-opening parabola with its peak at x = 1, y = 1. When rotated around the y-axis, it creates a rounded dome-like solid. The shell method captures this by summing shells of increasing radius from x = 0 to x = 2, each with a height matching the parabola at that x value.
What if the Axis Were x = −1?
If the axis of rotation were shifted to x = −1 instead of the y-axis, the radius of each shell would become (x − (−1)) = x + 1:
V = 2π ∫[0 to 2] (x + 1)(2x − x²) dx
This illustrates how the shell method adapts cleanly to offset axes — just update the radius expression.
Key Takeaways
- Always start by finding the intersection points to establish your limits
- Expand the integrand fully before integrating to avoid errors
- For two-curve problems, height = (top function) − (bottom function)
- For an offset axis, adjust the radius accordingly