Solución

Esta superficie tiene parametrización

r(x,θ)=⟨x,x2cos  θ,x2sen  θ⟩,0≤x≤b,0≤x<2π\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 tx=⟨1,2xcos  θ,2xsen  θ⟩\bold{t}_x = \lang 1, 2x cos\;\theta, 2x sen \;\theta\rang y tθ=⟨0,−x2sen  θ,−x2cos  θ⟩\bold{t}_{\theta} = \lang 0, −x^2sen \;\theta, −x^2cos \;\theta\rang. Por lo tanto,

tx×tθ=⟨2x3cos2θ+2x3sen2θ,−x2cos  θ,−x2senθ⟩=⟨2x3,−x2cos  θ,−x2senθ⟩\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

tx×tθ=4x6+x4cos2θ+x4sen2θ=4x6+x4=x24x2+1\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

∫0b∫0πx24x2+1=2π∫0bx24x2+1dx=2π[164(24x2+1(8x3+x)senh−1(2x))]0b=2π[164(24b2+1(8b3+b)senh−1(2b))]\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}