Surface Area obtained by rotating a curve?

  • Thread starter Flexer5
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  • #1
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1. Let the area of the surface obtained by rotating the curve x = 3e^(2y) from y=0 to y=1 about the y-axis



2. [tex]
S= \int_{y_1} ^{y_2} 2 \pi x ds \ where \ ds=\sqrt{1+ \left( \frac{dx}{dy} \right) ^2} dy
[/tex]




3. [tex]
S=2\pi \int _{0} ^{1} x\sqrt{1+\left(\frac{dx}{dy}\right)^2} dy
[/tex]
Let
[itex]
u=6e^{2y} \Rightarrow \frac{du}{dy}=12e^2y \Rightarrow \frac{du}{4}=3e^{2y}dy
[/itex]
[tex]
S=2\pi \int \frac{1}{4} \sqrt{1+u^2} du
[/tex]
I then got 2(pi) integral sec(theta)^3 d(theta) by substituting "u" for tan(theta)

I have been working on this problem for a while, and I am getting frustrated. What is the final result?
 
Last edited:

Answers and Replies

  • #2
Char. Limit
Gold Member
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All it takes is a u-substitution. Let u=6e^(2y)+1. Then du=12e^(2y) dy, and there's a nice factor of e^(2y) outside of your square root.
 
  • #3
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Hi,

this is correct, good work:

[tex]

S=2\pi \int \frac{1}{4} \sqrt{1+u^2} du

[/tex]

now u got basically an integral of I = sec(x)^3.
it's a tricky integral solving it is the art of integration :)
use integration by parts: u = sec(x); dv = sec(x)^2 dx
u get: I = uv - int( tan^2(x) sec(x) )

identity: tan^2(x) = sec^2(x) - 1
then you get the same integral on both sides of the equation but with opposite signs !

I = uv - int( sec^3(x) - sec(x) ) = uv - I + int( sec(x) )
I = 0.5* [ uv + int( sec(x) ) ]

here you go ! :)
 
  • #4
2
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Thank you guys for all your help.

Char. Limit made a great point, one that I overlooked. Here is what I did:

u == 6*E^(2y) + 1
du == 12*E^(2y)

integral_0^1 6e^(2y) pi sqrt(u) du
integral (pi sqrt(u))/2 du

After taking the integral I got: pi/(4 sqrt(u)).
I then plugged in for u getting: pi/(4 sqrt(6 e^(2 y)+1)) which I evaluated from 0 to 1, but I did not get the right answer! Not only that but the answer I got was negative.

Do I need to change the limits?
 
  • #5
Char. Limit
Gold Member
1,204
14
Thank you guys for all your help.

Char. Limit made a great point, one that I overlooked. Here is what I did:

u == 6*E^(2y) + 1
du == 12*E^(2y)

integral_0^1 6e^(2y) pi sqrt(u) du
integral (pi sqrt(u))/2 du

After taking the integral I got: pi/(4 sqrt(u)).
I then plugged in for u getting: pi/(4 sqrt(6 e^(2 y)+1)) which I evaluated from 0 to 1, but I did not get the right answer! Not only that but the answer I got was negative.

Do I need to change the limits?
Woops! Instead of integrating, you accidentally differentiated. The integral of sqrt(u) is (2/3)u^(3/2).
 
  • #6
275
0
I admit Char. Limit found a nice way to solve that integral, much better then messing with sec(x) :)
 

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