# Surface area of a curve

1. May 18, 2009

### compliant

1. The problem statement, all variables and given/known data
Find the surface area traced out when the curve 8y^2 = x^2(1-x^2) is revolved around the x-axis.

3. 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,
$$8y^2 = x^2(1-x^2)$$
$$y = \frac{\sqrt{x^2(1-x^2)}}{8}$$

Surface area = $$2\pi \int_{C} y dr$$
= $$4\pi \int^{1}_{0} y dr$$

$$dr = {\sqrt{1+ ({\frac{\sqrt{2x(1-2x^2)}}{8})^2}} {dx}$$
$$dr = {\frac{\sqrt{64+2x(1-2x^2)}}{8}} {dx}$$

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

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

Last edited: May 18, 2009
2. May 18, 2009

### tiny-tim

Hi compliant!

(have a square-root: √ and an integral: ∫ and try using the X2 tag just above the Reply box )
Nooo … using vertical slices, surface area = ∫2πydx times a factor to take account of the slope, which is … ?

3. May 18, 2009

### compliant

The factor is $$\sqrt{1+(dy/dx)^2}$$, which I'm sure I accounted for above.

$$dr = {\sqrt{1+ ({\frac{\sqrt{2x(1-2x^2)}}{8})^2}} {dx}$$

It might not be "dr", but I'm 99% sure that's the calculation of the factor.

4. May 18, 2009

### tiny-tim

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 = … ?

5. May 18, 2009

### compliant

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. May 18, 2009

### tiny-tim

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

7. May 18, 2009