Integration – Finding Volume

When the graph of \(y = f(x)\) is rotated around the x-axis from \(x=a\) to \(x=b\):

No +C is needed.

Always rewrite roots and reciprocals as powers before squaring:

• \(\sqrt{x} = x^{1/2}\Rightarrow [\sqrt{x}]^2 = x\)
• \(\frac{1}{x^3} = x^{-3}\Rightarrow [\frac{1}{x^3}]^2 = x^{-6}\)

Find the volume generated when \(y=x\) is rotated about the x-axis from \(x=0\) to \(x=2\).

\[ V = \pi \int_0^2 x^2 \, dx = \pi \left[\frac{x^3}{3}\right]_0^2 = \pi \cdot \frac{8}{3} = \frac{8\pi}{3}. \] Volume = \( \frac{8\pi}{3} \) cm\(^3\) (if units are in cm).

Find the volume generated when \(y=\sqrt{x}\) is rotated around the x-axis from \(x=1\) to \(x=4\).

Rewrite \( \sqrt{x} = x^{1/2} \Rightarrow [\sqrt{x}]^2 = x \) \[ V = \pi \int_1^4 x \, dx = \pi \left[\frac{x^2}{2}\right]_{1}^{4} = \pi\left(\frac{16}{2}-\frac{1}{2}\right) = \pi \cdot \frac{15}{2} = \frac{15\pi}{2}. \] Volume = \( \frac{15\pi}{2} \) units\(^3\).

Find the volume generated when \(y=\frac{2}{x^2}\) is rotated from \(x=1\) to \(x=3\).

Rewrite: \( \frac{2}{x^2} = 2x^{-2} \Rightarrow y^2 = 4x^{-4}\) \[ V = \pi \int_1^3 4x^{-4}\,dx = 4\pi \left[\frac{x^{-3}}{-3}\right]_1^3 = -\frac{4\pi}{3}\left[\frac{1}{3^3} - 1 \right] = -\frac{4\pi}{3}\left(\frac{1}{27}-1\right) = \frac{104\pi}{81}. \] Volume = \( \frac{104\pi}{81} \) units\(^3\).

• Always square the function before integrating.
• Never include +C for volume.
• Rewrite roots and reciprocals as powers before squaring.
• If the curve touches the x-axis, check if the lower limit is 0 (works fine in Year 12 only for powers where \(n > -1\)).
• Units in final answer are cubed: \( \text{units}^3 \).