Let z=f(x), x in[a,b], be a curve in the xz -plane (y=0)

Let $z=f(x)$, $x\in[a,b]$, be a curve in the $xz$-plane ($y=0$), with $f$ continuously differentiable. Derive the formula for the area of the surface generated by rotating this curve about the $Oz$ axis, as a double integral (and reduce it to a single integral).

Step 1 · Parametrization

$$ \mathbf r(x,\theta) = \big(x\cos\theta,\ x\sin\theta,\ f(x)\big), \qquad x\in[a,b],\ \theta\in[0,2\pi]. $$

Step 2 · Cross product and its magnitude

$$ \mathbf r_x = \big(\cos\theta,\ \sin\theta,\ f'(x)\big), \qquad \mathbf r_\theta = \big(-x\sin\theta,\ x\cos\theta,\ 0\big). $$

$$ \mathbf r_x\times\mathbf r_\theta = \big(-x f'(x)\cos\theta,\ -xf'(x)\sin\theta,\ x\big). $$

$$ \|\mathbf r_x\times\mathbf r_\theta\| = \sqrt{x^2\big[f'(x)\big]^2\cos^2\theta + x^2\big[f'(x)\big]^2\sin^2\theta + x^2} = x\sqrt{1+\big[f'(x)\big]^2}, $$

(taking $x\ge0$ on $[a,b]$, so $|x|=x$).

Step 3 · Reduce the double integral to a single integral

$$ S=\int_0^{2\pi}\int_a^b x\sqrt{1+\big[f'(x)\big]^2}\,dx\,d\theta = 2\pi\int_a^b x\sqrt{1+\big[f'(x)\big]^2}\,dx, $$

since the integrand is independent of $\theta$.

Additional Notes

• This is the standard, well-known formula for the area of a surface of revolution generated by rotating $z=f(x)$ about the $Oz$ axis; the source's derivation follows essentially the same route and is correct.
• If the generating curve is instead given parametrically as $x=x(t),\,z=z(t)$, $t\in[t_1,t_2]$, the same argument gives the equivalent form $S=2\pi\displaystyle\int_{t_1}^{t_2}x(t)\sqrt{\big[x'(t)\big]^2+\big[z'(t)\big]^2}\,dt$.
• As a sanity check, applying the formula to a sphere of radius $R$ (upper half, $f(x)=\sqrt{R^2-x^2}$, $x\in[0,R]$) recovers the classical hemisphere area $2\pi R^2$: here $x\sqrt{1+f'(x)^2}=x\cdot\dfrac R{\sqrt{R^2-x^2}}$, and $2\pi\int_0^R\dfrac{Rx}{\sqrt{R^2-x^2}}\,dx = 2\pi R^2$.