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}