How do you find the area of the surface generated by rotating the curve about the y-axis #x=t^2, y=1/3t^3, 0

Question

How do you find the area of the surface generated by rotating the curve about the y-axis #x=t^2, y=1/3t^3, 0<=t<=3#?

Samantha Adkison · Accepted Answer

For infinitesimal arc length, #ds#:

#ds = sqrt(dx^2 + dy^2 )#

#:. ds = sqrt(4t^2 + t^4 ) \ dt = t sqrt(4 + t^2 ) \ dt#

Spinning round the y-axis, creating infinitesimal surface area #dS#, where:

#= 2 pi \ t^3 \ sqrt(4 + t^2 ) \ dt#

#:. S =2 pi int_0^3 \ dt qquad t^3 sqrt(4 + t^2 ) #

#= (2 pi)/15 (64 + 247 sqrt13)#

I have skipped the mechanical integration steps. you can start with a sub:

#:. S =2 pi int_4^13 \ (du)/(2t) qquad t^3 \ sqrtu#

#= pi int_4^13 \ du qquad (u - 4) \ sqrtu#

#= pi int_4^13 \ du qquad u^(3/2) - 4u^(1/2) #

Which is trivial but protracted

Christopher Agresta · Answer

undefined To find the area of the surface generated by rotating the curve about the y-axis $ x = t^2 $, $ y = \frac{1}{3}t^3 $ for $ 0 \leq t \leq 3 $, you can use the formula for surface area of revolution:

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

First, express $ x $ in terms of $ y $ to get $ x = (3y)^{2/3} $. Then, compute $ \frac{dy}{dx} $ and substitute the expressions into the formula to calculate the integral. This will give you the area of the surface generated by rotating the curve about the y-axis.

How do you find the area of the surface generated by rotating the curve about the y-axis #x=t^2, y=1/3t^3, 0&lt;=t&lt;=3#?

How do you find the area of the surface generated by rotating the curve about the y-axis #x=t^2, y=1/3t^3, 0<=t<=3#?