Expert: Alon Mandes - 9/14/2008

Question
Find the area of the surface obtained by rotating the curve about the x-axis.
y = x^3
0 ≤ x ≤ 5

and how is it different from solving one like:

The given curve is rotated about the y-axis. Find the area of the resulting surface.
y = root3(x)
4 ≤ y ≤ 5


thank you so much if you can help, my teacher isn't the greatest at explaining these.

Answer
Hello Carrie, let's solve this exercise step by step, shall we?
OK, 1st let's write the formula of surface area of rotating :

If the rotation is performing about the x-axis, then the 3-dimensional object obtained will be horizontal, & we will use x as
variable & y=f(x). & the formula for the area will be :  
A= ∫2*π*y(x)*sqrt[1+y'(x)^2]dx. (sqrt means root)
Here y(x)=x^3, so y'(x)=3x^2 & y'(x)^2=9x^4.
Now we can calculate the integral. Notice that the boundaries of
our integral is from 0 to 5. Ok them , let's calculate :
∫2*π*x^3*sqrt[1+9x^4]dx.
Carrie I will leave it to you from here to practice this integral,
it's easy, use the substitution method.


If the rotation is performing about the y-axis, then the 3-dimensional object obtained will be vertical, & the formula for
the area will be :  
A= ∫2*π*x*sqrt[1+y'(x)^2]dx. Notice 2 things :
1st : This time we used x instead of f(x) in the integral
2nd : The boundaries are not from 4 to 5..No no, we have to calculate them.
Let's calculate the derivative : y=x^(1/3) --> y'=(1/3)x^(-2/3).
So, in our case we have ∫2*π*x*sqrt[1+(1/9)x^(-4/3)]dx.
Now Carrie, we have to calculate the limits of the integral :
if y goes from 4 to 5, then it's logical to claim that x goes
from 4^3 to 5^3, right ? because y=x^(1/3) so x=y^3.
Hence, our boundaries are : x goes from 64 to 125.
& I will leave the calculation of the integral to you as an
exercise.
Any problems Carrie, I'm here.

Alon.