Surface Area of Revolution

Rotating $y = f(x)$ around the x-axis creates a surface. Its area is:

$SA = 2pi int_a^b f(x) cdot sqrt{1 + [f'(x)]^2} , dx$

Why It Works

Each strip of width $dx$ wraps into a ribbon of radius $f(x)$ and slant height $sqrt{1+f'^2},dx$ . The ribbon's area is $2pi cdot f(x) cdot sqrt{1+f'^2},dx$ .

Classic Examples

Cylinder $f(x) = r$ on $[0, L]$ : $f' = 0$ , so $SA = 2pi r L$

Cone $f(x) = x$ on $[0, 1]$ : $f' = 1$ , so:

ight]_0^1 = pisqrt{2} approx 4.4429$$ ### Numerical Approach ```c #define PI 3.14159265358979 for each midpoint x: double fp = (f(x+h) - f(x-h)) / (2*h); sum += f(x) * my_sqrt(1.0 + fp * fp); return 2.0 * PI * sum * dx; ``` ### Your Task Implement `double surface_area(double (*f)(double), double a, double b, int n, double h)`.

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