My Math Forum Surface area of a prolate spheroid

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 February 10th, 2017, 08:11 AM #1 Newbie   Joined: Feb 2017 From: Finland Posts: 2 Thanks: 0 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!
 February 11th, 2017, 09:12 AM #2 Member   Joined: Feb 2015 From: Southwest Posts: 96 Thanks: 24 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.
 February 26th, 2017, 01:13 AM #3 Newbie   Joined: Feb 2017 From: Finland Posts: 2 Thanks: 0 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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### proof of surface area of a prolate spheriod

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