Find the area of the surface obtained by rotating the curve of parametric equations:
x=3t−3/3(t^3) y=3t^2 0

Question

Find the area of the surface obtained by rotating the curve of parametric equations:
x=3t−3/3(t^3)   y=3t^2     0<t<1
about the x - axis.

anonymous · Answer

So I set up my integral like so : $$2\pi \int\limits_{0}^{1}(3t-t^3)\sqrt{(-3t^2+3)^2+(6t)^2}$$

anonymous · Answer

After evaluating it I get an answer of 11pi but that is incorrect. Any help would be really appreciated!

accessdenied · Answer

i agree with most of it:  what was your reasoning for the choice of using x=3t - t^3   ds ?

anonymous · Answer

Because it's being rotated about the x-axis

accessdenied · Answer

hm..  but if we revolve it around the x-axis:

accessdenied · Answer

|dw:1395605096452:dw|
this radius is the y-value, correct?

anonymous · Answer

Yes

accessdenied · Answer

basically, the surface area is being defined as the sum of each circumference of a circle:
  2 pi r
which we add up through all small lengths ds along the curve  (which we convert to dt)
we just said that radius was y?
  integral of 2 pi * y  * ds
rather than 2 pi x

accessdenied · Answer

so i think if you use y = 3t^2 rather than x = 3t - t^3, you should get the correct integral and correct answer (looking at http://tutorial.math.lamar.edu/Classes/CalcII/ParaSurfaceArea.aspx it is also the given formula for rotation about x-axis)

anonymous · Answer

Ohhh okay I see. I got the right answer now! Thanks so much for the help.

accessdenied · Answer

glad to help!  :)