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?

Question

Ezra Adams · Accepted Answer

#=489392.457# Cubic units#

Parametric Equations

#x=3t-1#

#thereforet=(x+1)/3##color(green)(rarr(1)#

#y=t^2-2t+2#

Substitute in #y#

#y=(x+1)^2/9-2xx(x+1)/3+2#

When

#tin[2,3]##rarr##x in[5,8]# #color(blue) (" from "##color(green)((1)#

Determining the Volume of a Solid of Revolution

#V=piint_5^8((x+1)^2/9-2xx(x+1)/3+2)^2#

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

#=489392.457# Cubic units#

Adam Adkins · Answer

undefined To find the surface area of the solid formed by revolving the curve $ f(t) = (3t-1, t^2-2t+2) $ around the x-axis from $ t = 2 $ to $ t = 3 $, you can use the formula for the surface area of revolution:

$$ S = \int_{a}^{b} 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dt $$

Where:
- $ a $ and $ b $ are the limits of integration (in this case, $ a = 2 $ and $ b = 3 $)
- $ y $ represents the $ y $-coordinate of the curve $ f(t) $
- $ \frac{dy}{dx} $ represents the derivative of $ y $ with respect to $ x $

First, find the derivative $ \frac{dy}{dx} $ by differentiating the equation $ y = t^2-2t+2 $ with respect to $ t $. Then substitute into the formula and integrate over the given interval to find the surface area.