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February 10th, 2017, 08:11 AM   #1
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Surface area of a prolate spheroid

Hello guys!
I'm trying to calculate the surface area of a prolate spheroid and I found this formula from Prolate Spheroid -- from Wolfram MathWorld
image.jpg

However, I haven't been able to proof that r(z)(1+r'(z)^2)^1/2 = a(1+((a-c)(a+c)z^2)/c^4)^1/2

I know the left side is just derivated and solved algebraically but I still don't manage to proof it.

I would be very grateful if someone could help me!
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February 11th, 2017, 09:12 AM   #2
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I got

$\displaystyle r'(z)=\frac{-az}{c^2\sqrt{1-(\frac{z}{c})^2}}$
$\displaystyle r'(z)^2=\frac{a^2z^2}{c^4-c^2z^2}$

Then I tried inserting all that into the integrand, but I got lost in it.
I'm hoping my failed work triggers success in yours.

$\displaystyle r(z)\sqrt{1+r'(z)^2}=a\sqrt{(1-\frac{z^2}{c^2})(1+\frac{a^2z^2}{c^4-c^2z^2})}$

$\displaystyle r(z)\sqrt{1+r'(z)^2}=a\sqrt{1+\frac{a^2z^2}{c^4-c^2z^2}-\frac{z^2}{c^2}-\frac{a^2z^4}{c^6-c^4z^2}}$

$\displaystyle r(z)\sqrt{1+r'(z)^2}=a\sqrt{1+\frac{c^2a^2z^2}{c^6-c^4z^2}-\frac{c^4z^2-c^4z^4}{c^6-c^4z^2}-\frac{a^2z^4}{c^6-c^4z^2}}$

This is where I got stuck.
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February 26th, 2017, 01:13 AM   #3
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Thanks for your help! Later on I realized if I take common factors and cancell some stuff I will get the correct form
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