# What is the surface area of the solid created by revolving f(t) = ( 3t-1, t^2-2t+2), t in [2,3] around the x-axis?

Apr 23, 2018

$= 489392.457$ Cubic units

#### Explanation:

$x = 3 t - 1$

$\therefore t = \frac{x + 1}{3}$color(green)(rarr(1)

$y = {t}^{2} - 2 t + 2$

Substitute in $y$

$y = {\left(x + 1\right)}^{2} / 9 - 2 \times \frac{x + 1}{3} + 2$

When

$t \in \left[2 , 3\right]$$\rightarrow$$x \in \left[5 , 8\right]$ color(blue) (" from "color(green)((1)

$V = \pi {\int}_{5}^{8} {\left({\left(x + 1\right)}^{2} / 9 - 2 \times \frac{x + 1}{3} + 2\right)}^{2}$

For this Integration, You will use color(blue)("The Binomial Theorem and The Power Rule" 

$= 489392.457$ Cubic units