Math · Calculus II · Worked example

Find a volume with the washer method

The region between y = x and y = x² is revolved about the x-axis. Find the volume of the solid.

Find the limits

The curves meet where x² = x, at x = 0 and x = 1. Between them the line y = x lies above the parabola.

\begin{array}{l} x^{2} = x \\ x = 0 or x = 1 \end{array}

Outer and inner radii

A slice at x is a washer. Its outer radius is the line’s height, R = x, and its inner radius is the parabola’s, r = x².

Integrate the washer areas

Subtract the areas: π(R² − r²) = π(x² − x⁴).

V = \int_{0}^{1} π (x^{2} - x^{4}) d x = π (\frac{1}{3} - \frac{1}{5}) = \frac{2 π}{15}

\begin{array}{c} V = \int_{0}^{1} π (x^{2} - x^{4}) d x \\ = π (\frac{1}{3} - \frac{1}{5}) = \frac{2 π}{15} \end{array}

Result

V = 2π/15 ≈ 0.419 cubic units.

Your turn

Revolve the region between y = 2x and y = x² about the x-axis. Find the volume.

Show the answer and explanation

64π/15 ≈ 13.4.

The curves meet at x = 0 and x = 2, with the line on top. V = ∫₀² π((2x)² − (x²)²) dx = π(32/3 − 32/5) = 64π/15.

\begin{array}{l} x^{2} = 2 x \\ x = 0 or x = 2 \\ \int_{0}^{2} π (4 x^{2} - x^{4}) d x = π (\frac{32}{3} - \frac{32}{5}) = \frac{64 π}{15} \end{array}

\begin{array}{l} x^{2} = 2 x \\ x = 0 or x = 2 \\ \int_{0}^{2} π (4 x^{2} - x^{4}) d x \\ = π (\frac{32}{3} - \frac{32}{5}) = \frac{64 π}{15} \end{array}

Keep exploring

In Slopes, sums & signed area, enter π(x − x²)², the result of subtracting the radii before squaring. It gives π/30 ≈ 0.105, only a quarter of the true volume.

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