Question

Consider the parametric equations below.

Answer 1

The area of the surface is

$\displaystyle2\pi\int_0^{\pi/6}|y(t)|\sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt$

$t\cos t\ge0$ for all $0\le t\le\dfrac\pi6$, and

$\dfrac{\mathrm dx}{\mathrm dt}=t\cos t+\sin t$

$\dfrac{\mathrm dy}{\mathrm dt}=-t\sin t+\cos t$

$\implies\left(\dfrac{\mathrm dx}{\mathrm dt}\right)^2+\left(\dfrac{\mathrm dy}{\mathrm dt}\right)^2=(t\cos t+\sin t)^2+(t\sin t-\cos t)^2=t^2+1$

The integral is then

$\displaystyle2\pi\int_0^{\pi/6}t\cos t\sqrt{t^2+1}\,\mathrm dt\approx0.8547$