4. Volumes of Revolution

Video coming soon2 worked examples

Theory

Rotating a curve about an axis sweeps out a solid whose volume is found by integrating the area of circular cross-sections. About the $x$ -axis, between $x = a$ and $x = b$ :

V = \pi \int_a^b y^2\,dx

About the $y$ -axis, between $y = c$ and $y = d$ : $V = \pi \int_c^d x^2\,dy$ .

The Golden Rule: square the radius (the function), integrate, and multiply by $\pi$ ; rotating about the $x$ -axis uses $\int y^2\,dx$ , about the $y$ -axis uses $\int x^2\,dy$ .

⚠️ Common Examiner Traps

Forgetting to square: the integrand is $y^2$ (or $x^2$ ), not $y$ .
Wrong variable: rotation about the $y$ -axis integrates with respect to $y$ .
Losing the $\pi$ : the whole result is multiplied by $\pi$ .

Worked examples

Example 1

The curve $y = \sqrt{x}$ is rotated about the $x$ -axis between $x = 0$ and $x = 4$ . Find the volume generated.

Step 1: Apply the formula with $y^2 = (\sqrt{x})^2 = x$ :

V = \pi \int_0^4 x\,dx

Step 2: Integrate and evaluate:

= \pi\left[\frac{x^2}{2}\right]_0^4 = \pi\left(\frac{16}{2}\right) = 8\pi

Example 2

The curve $y = x^2$ is rotated about the $y$ -axis between $y = 0$ and $y = 9$ . Find the volume generated.

Step 1: Rotating about the $y$ -axis uses $\int x^2\,dy$ . Here $y = x^2$ , so $x^2 = y$ :

V = \pi \int_0^9 y\,dy

Step 2: Integrate and evaluate:

= \pi\left[\frac{y^2}{2}\right]_0^9 = \pi\left(\frac{81}{2}\right) = \frac{81\pi}{2}