Math · Calculus II · Worked example

Find a volume with the shell method

The region between y = 2x − x² and the x-axis is revolved about the y-axis. Find the volume of the solid.

Why shells

Washers would need x in terms of y, which means solving y = 2x − x² for x. Vertical strips avoid that: turned about the y-axis, each strip sweeps out a shell.

Find the limits

The parabola meets the x-axis at x = 0 and x = 2.

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

Radius and height

The strip at x is x units from the axis and 2x − x² tall, so the shell’s volume is about 2πx(2x − x²) dx.

Integrate the shells

Expand before integrating: x(2x − x²) = 2x² − x³.

V = \int_{0}^{2} 2 π (2 x^{2} - x^{3}) d x = 2 π (\frac{16}{3} - 4) = \frac{8 π}{3}

\begin{array}{c} V = \int_{0}^{2} 2 π (2 x^{2} - x^{3}) d x \\ = 2 π (\frac{16}{3} - 4) = \frac{8 π}{3} \end{array}

Result

V = 8π/3 ≈ 8.38 cubic units.

Your turn

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

Show the answer and explanation

π/6 ≈ 0.524.

Shells of radius x and height x − x² from x = 0 to x = 1: V = ∫₀¹ 2πx(x − x²) dx = 2π(1/3 − 1/4) = π/6.

\begin{array}{l} x^{2} = x \\ x = 0 or x = 1 \\ \int_{0}^{1} 2 π (x^{2} - x^{3}) d x = 2 π (\frac{1}{3} - \frac{1}{4}) = \frac{π}{6} \end{array}

\begin{array}{l} x^{2} = x \\ x = 0 or x = 1 \\ \int_{0}^{1} 2 π (x^{2} - x^{3}) d x \\ = 2 π (\frac{1}{3} - \frac{1}{4}) = \frac{π}{6} \end{array}

Keep exploring

In Slopes, sums & signed area, turn the same region about the x-axis instead, with disks: enter π(2x − x²)². The volume is 16π/15 ≈ 3.35, a different solid from the same region.

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