• Support PF! Buy your school textbooks, materials and every day products Here!

Surface area of a curve around the x-axis

  • Thread starter compliant
  • Start date
  • #1
45
0

Homework Statement


Find the surface area traced out when the curve 8y^2 = x^2(1-x^2) is revolved around the x-axis.

The Attempt at a Solution


x-axis means y = 0
When y = 0, x = 0, -1 or 1.
Since this curve is "the infinity symbol", the curve has symmetry at x = 0.

Isolating y,
[tex]8y^2 = x^2(1-x^2)[/tex]
[tex]y = \frac{\sqrt{x^2(1-x^2)}}{8}[/tex]

Surface area = [tex]2\pi \int_{C} y dr [/tex]
= [tex]4\pi \int^{1}_{0} y dr [/tex]

[tex] dr = {\sqrt{1+ ({\frac{\sqrt{2x(1-2x^2)}}{8})^2}} {dx}[/tex]
[tex] dr = {\frac{\sqrt{64+2x(1-2x^2)}}{8}} {dx}[/tex]

Surface area = [tex]{\frac{\pi}{8}} \int^{1}_{0} {\sqrt{x^2(1-x^2)(64+2x(1-2x^2))}{dx}[/tex]


And.......I have no idea how to proceed with the integral.
 
Last edited:

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
249
Hi compliant! :smile:

(have a square-root: √ and an integral: ∫ and try using the X2 tag just above the Reply box :wink:)
Find the surface area traced out when the curve 8y^2 = x^2(1-x^2) is revolved around the x-axis.
Surface area = [tex]2\pi \int_{C} y dr [/tex] …
Nooo … using vertical slices, surface area = ∫2πydx times a factor to take account of the slope, which is … ? :smile:
 
  • #3
45
0
The factor is [tex]\sqrt{1+(dy/dx)^2}[/tex], which I'm sure I accounted for above.

[tex] dr = {\sqrt{1+ ({\frac{\sqrt{2x(1-2x^2)}}{8})^2}} {dx}[/tex]

It might not be "dr", but I'm 99% sure that's the calculation of the factor.
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
249
oh i see … yes that's right … i didn't recognise it from your:
[tex] dr = {\sqrt{1+ ({\frac{\sqrt{2x(1-2x^2)}}{8})^2}} {dx}[/tex]
start again … if y = (1/√8)x√(1 - x2), then dy/dx = … ? :smile:
 
  • #5
45
0
oh i see … yes that's right … i didn't recognise it from your:


start again … if y = (1/√8)x√(1 - x2), then dy/dx = … ?
Ah, I must've forgotten to chain rule it.



y = (1/√8)√(x2 - x4)
dy/dx = (1/2√8)(x2 - x4)-1/2(2x - 4x3)
dy/dx = (1/√8)(x2 - x4)-1/2(x - 2x3)
 
  • #6
tiny-tim
Science Advisor
Homework Helper
25,832
249
dy/dx = (1/√8)(x2 - x4)-1/2(x - 2x3)
yes, that looks right … (you had a factor 2 earlier) …

now square it and add the 1, and I think it'll be a perfect square on the top :wink:
 
  • #7
45
0
Ok, got the answer. Thanks.
 

Related Threads for: Surface area of a curve around the x-axis

Replies
1
Views
957
Replies
6
Views
5K
  • Last Post
Replies
2
Views
868
Replies
2
Views
4K
Replies
5
Views
2K
Replies
6
Views
4K
Replies
1
Views
2K
Top