Surface Area with Parametric Equations — Question 4

Question 4

Problem

Set up, but do not evaluate, the area formed by rotating $x=t^2$, $y=t^3$, $0\le t\le1$, about the $y$-axis.

Question 4 – Solution

Solution

1. Differentiate the coordinates: $$x'(t)=2t, \qquad y'(t)=3t^2.$$

2. Compute the arc-length element: $$\begin{aligned} ds&=\sqrt{(2t)^2+(3t^2)^2}\,dt\\ &=\sqrt{4t^2+9t^4}\,dt =|t|\sqrt{4+9t^2}\,dt. \end{aligned}$$ Since $0\le t\le1$, this becomes $ds=t\sqrt{4+9t^2}\,dt$.

3. Rotation is about the $y$-axis, so the radius is the distance $x=t^2$.

4. Therefore, without evaluating, $$\begin{aligned} S&=2\pi\int_0^1x(t)\,ds\\ &=\boxed{2\pi\int_0^1t^3\sqrt{4+9t^2}\,dt}. \end{aligned}$$