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MathCalculus

Shell Method Formula: Volume of y = 3x − x² About the y-Axis

Using shells of radius x and height 3x − x² gives V = 2π∫₀²x(3x − x²) dx = 8π cubic units.

Question

Define RR as the region bounded above by f(x)=3x−x2f(x)=3x-x^2 and below by the xx-axis over the interval [0,2][0,2].

Find the volume of the solid of revolution formed by revolving RR around the yy-axis.

Answer

A vertical strip at xx forms a cylindrical shell with

  • radius r=xr=x,
  • height h=3x−x2h=3x-x^2,
  • thickness dxdx.

The shell method formula is

V=2π∫ab(radius)(height) dx.V=2\pi\int_a^b (\text{radius})(\text{height})\,dx.

Here,

V=2π∫02x(3x−x2) dx=2π∫02(3x2−x3) dx=2π[x3−x44]02=2π(8−4)=8π.\begin{aligned} V &=2\pi\int_0^2 x(3x-x^2)\,dx\\ &=2\pi\int_0^2(3x^2-x^3)\,dx\\ &=2\pi\left[x^3-\frac{x^4}{4}\right]_0^2\\ &=2\pi(8-4)\\ &=\boxed{8\pi}. \end{aligned}

The volume is 8π8\pi cubic units.

Evidence boundary

The computation uses the entire region under y = 3x − x² on [0, 2] and rotation about the y-axis. The result is exact in cubic units because the prompt gives no physical unit.

Sources

These references support the concepts and methods used in the explanation above.

Shell Method Formula: Volume Worked Example | Verla