# 11.2 #60

Surface area of the curve $x=3t-t^3, y=3t^2, 0\leq\,t\,\leq\,1$

1. Find dx/dt and dy/dt: $3-3t^2$ and $6t$, respectively
2. Square them: $9-18t^{2}+9t^{4}$ and $36t^{2}$, respectively
3. $SA=\int_{0}^{1}2\pi\,3t^2\sqrt{9t^{4}-18t^{2}+9+36t^{2}}dt$
4. Factor out the root's 9 and move coefficients out: $18\pi\,\int_{0}^{1}t^{2}\sqrt{t^{4}+2t^{2}+1}dt$
5. Note the root contains a square: $(t^{2}+1)^{2}$.
6. Thus: $18\pi\,\int_{0}^{1}t^{2}(t^{2}+1)dt$
7. $18\pi\,\int_{0}^{1}\,(t^{4}+t^{2})dt=18\pi\,[\frac{1}{5}t^{5}+\frac{1}{3}t^{3}]=18\pi\,(\frac{1}{5}+\frac{1}{3})=\frac{144\pi}{15}$