Surface Area — Question 9

Question 9

Parametric curve: $x=t^2$, $y=1+t^3/3$, for $0\le t\le1$.
Axis of rotation: the $x$-axis.
Task: Set up, but do not evaluate, the surface-area integral.

Question 9 – Solution

Step 1: Differentiate both coordinates. $$\frac{dx}{dt}=2t,\qquad\frac{dy}{dt}=t^2.$$ Step 2: Find the parametric arc-length factor. $$\begin{align*} ds&=\sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2}\,dt\\ &=\sqrt{4t^2+t^4}\,dt\\ &=t\sqrt{4+t^2}\,dt \end{align*}$$ because $t\ge0$.

Step 3: Use radius $y=1+t^3/3$. $$\begin{align*} S&=2\pi\int_0^1y\,ds\\ &=2\pi\int_0^1\left(1+\frac{t^3}{3}\right) t\sqrt{4+t^2}\,dt\\ &=2\pi\int_0^1\left(t+\frac{t^4}{3}\right) \sqrt{4+t^2}\,dt. \end{align*}$$ $$\boxed{S=2\pi\int_0^1\left(t+\frac{t^4}{3}\right)\sqrt{4+t^2}\,dt}$$