Surface area of a curve around the x-axis

[compliant]

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,
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.

 

[tiny-tim]

Hi compliant!

(have a square-root: √ and an integral: ∫ and try using the X² tag just above the Reply box )

Find the surface area traced out when the curve 8y^2 = x^2(1-x^2) is revolved around the x-axis.
Surface area = 2\pi \int_{C} y dr …

Nooo … using vertical slices, surface area = ∫2πydx times a factor to take account of the slope, which is … ?

 

[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.

 

[tiny-tim]

oh i see … yes that's right … i didn't recognise it from your:

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

start again … if y = (1/√8)x√(1 - x²), then dy/dx = … ?

 

[compliant]

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)

 

[tiny-tim]

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

 

[compliant]

Ok, got the answer. Thanks.