# How do you find the area of the surface generated by rotating the curve about the y-axis x=t+1, y=1/2t^2+t, 0<=t<=2?

Jun 25, 2018

$= \frac{2 \pi}{3} \left(10 \sqrt{10} - 2 \sqrt{2}\right) \text{ sq units}$

#### Explanation:

$\boldsymbol{r} \left(t\right) = \left\langlet + 1 , \frac{1}{2} t \left(t + 2\right)\right\rangle , q \quad 0 \le t \le 2$

$\mathrm{ds} = \sqrt{{\dot{x}}^{2} + {\dot{y}}^{2}} \setminus \mathrm{dt}$

$\mathrm{dS} = \mathrm{ds} \cdot 2 \pi x = 2 \pi \left(t + 1\right) \sqrt{{\left(1\right)}^{2} + {\left(t + 1\right)}^{2}} \setminus \mathrm{dt}$

$= 2 \pi \left(t + 1\right) \sqrt{{t}^{2} + 2 t + 2} \setminus \mathrm{dt}$

$S = 2 \pi {\int}_{0}^{2} \mathrm{dt} q \quad \left(t + 1\right) \sqrt{{t}^{2} + 2 t + 2}$

$= 2 \pi {\int}_{0}^{2} \mathrm{dt} q \quad \frac{d}{\mathrm{dt}} \left(\frac{1}{3} {\left({t}^{2} + 2 t + 2\right)}^{\frac{3}{2}}\right)$

$= \frac{2 \pi}{3} {\left[{\left({t}^{2} + 2 t + 2\right)}^{\frac{3}{2}}\right]}_{0}^{2}$

$= \frac{2 \pi}{3} \left({10}^{\frac{3}{2}} - {2}^{\frac{3}{2}}\right)$

$= \frac{2 \pi}{3} \left(10 \sqrt{10} - 2 \sqrt{2}\right) \approx 60.3 \text{ sq units}$