Esta superficie tiene parametrización
$$\bold{r}(x, \theta) = \lang x, x^2cos\; \theta, x^2 sen \;\theta\rang , 0 \le x \le b, 0 \le x \lt 2\pi$$Los vectores tangentes son $\bold{t}_x = \lang 1, 2x cos\;\theta, 2x sen \;\theta\rang$ y $\bold{t}_{\theta} = \lang 0, −x^2sen \;\theta, −x^2cos \;\theta\rang$. Por lo tanto,
$$\begin{aligned} \bold{t}_x \times \bold{t}_{\theta} &= \lang 2x^3cos^2\theta + 2x^3 sen^2\theta, −x^2cos\;\theta, −x^2 sen \theta\rang\\ &= \lang 2x^3, −x^2cos\;\theta, −x^2sen \theta\rang \end{aligned}$$y
$$\begin{aligned} \bold{t}_x \times \bold{t}_{\theta} &= \sqrt{4x^6 + x^4cos^2\theta + x^4sen^2\theta}\\ &= \sqrt{4x^6 + x^4}\\ &= x^2\sqrt{4x^2+1} \end{aligned}$$El área de la superficie de revolución es
$$\begin{aligned} \int_0^b\int_0^{\pi} x^2\sqrt{4x^2+1} &= 2\pi \int_0^b x^2\sqrt{4x^2+1}dx\\ &= 2\pi\bigg[\frac{1}{64}\bigg(2\sqrt{4x^2 + 1}\big(8x^3+x\big)senh^{-1}(2x)\bigg)\bigg]_0^b\\ &= 2\pi\bigg[\frac{1}{64}\bigg(2\sqrt{4b^2 + 1}\big(8b^3+b\big)senh^{-1}(2b)\bigg)\bigg] \end{aligned}$$