## Calculus: Early Transcendentals 8th Edition

$$x= t^{2}-t^{3}, \quad y= t+t^{4} , \quad 0 \leq t \leq 1$$ The surface area of the curve, which is obtained by rotating the given curve about the x-axis, is $$S=\int 2 \pi y d s=\int_{0}^{1} 2 \pi\left(t+t^{4}\right) \sqrt{16 t^{6}+9 t^{4}-4 t^{3}+4 t^{2}+1} d t \approx 12.7176$$
$$x= t^{2}-t^{3}, \quad y= t+t^{4} , \quad 0 \leq t \leq 1$$ then $$d x / d t=2 t-3 t^{2} , \quad d y / d t =1+4 t^{3}$$ so \begin{aligned}\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2} &=\left(2 t-3 t^{2}\right)^{2}+\left(1+4 t^{3}\right)^{2} \\ &=4 t^{2}-12 t^{3}+9 t^{4}+1+8 t^{3}+16 t^{6}, \end{aligned} thus the surface area of the curve, which is obtained by rotating the given curve about the x-axis, is $$S=\int 2 \pi y d s=\int_{0}^{1} 2 \pi\left(t+t^{4}\right) \sqrt{16 t^{6}+9 t^{4}-4 t^{3}+4 t^{2}+1} d t \approx 12.7176$$