Volume of Revolution

Volume of Revolution

Rotating a curve $y = f(x)$ around the x-axis sweeps out a solid. We can find its volume using the disk method.

Disk Method

At each $x$, the cross-section is a disk with radius $f(x)$ and area $\pi [f(x)]^2$. Integrating these areas:

$V = \pi \int_a^b [f(x)]^2\, dx$

Washer Method

Rotating the region between two curves $f(x)$ (outer) and $g(x)$ (inner):

$V = \pi \int_a^b \left([f(x)]^2 - [g(x)]^2\right) dx$

Each cross-section is a washer (disk with hole).

Classic Examples

Cylinder: rotate $f(x) = r$ (constant) on $[0, h]$:

$V = \pi r^2 h$

Cone: rotate $f(x) = \frac{rx}{h}$ (line) on $[0, h]$:

$V = \pi \int_0^h \left(\frac{rx}{h}\right)^2 dx = \frac{\pi r^2}{h^2} \cdot \frac{h^3}{3} = \frac{\pi r^2 h}{3}$

Sphere: rotate $f(x) = \sqrt{r^2 - x^2}$ on $[-r, r]$:

$V = \pi \int_{-r}^{r} (r^2 - x^2)\, dx = \frac{4}{3}\pi r^3$

Numerical Implementation

Use the midpoint rule:

#define PI 3.14159265358979
for each midpoint x:
    sum += f(x) * f(x)
return PI * sum * h

Your Task

Implement double volume_disk(double (*f)(double), double a, double b, int n) that computes $\pi \int_a^b [f(x)]^2\, dx$ using the midpoint rule.