Calculating Area of Ellipsoid: Surfaces of Revolution

[Lanza52]

Consider the ellipse:

(\frac{x}{2})^2 + y^2 = 1


We rotate this ellipse about the x-axis to form a surface known as ellipsoid. Determine the area of this surface.

Start off by solving for y.

y = \sqrt{1-\frac{x^2}{4}}

Then find the derivative.

y' = \frac{-x}{2\sqrt{4-x^2}}

Then plug into the formula for surface of revolution.

S = \int 2\pi y \sqrt{1+(\frac{dy}{dx})^2} dx

2 \pi \int \sqrt{1-\frac{x^2}{4}} \sqrt{1+(\frac{-x}{2\sqrt{4-x^2}})^2} dx


Plenty of simplifications later yields

\frac{\pi}{2} \int \sqrt{16-3x^2} dx

Now I haven't found that anti-derivative yet but just looking at it tells you its going to be extremely ugly. But all I remember my professor showing us in class was beautiful little problems that come out to \int 4x dx or something simple like that. So that makes me think I'm wrong.

Any help?

 

[rock.freak667]

Well I found a formula for it but I am not really sure how to show the working for it. But here is the formula

\int \sqrt{a^2-x^2} dx =\frac{1}{2}(x\sqrt{a^2-x^2}+a^2sin^{-1}(\frac{x}{a}))+ c

 

[Lanza52]

Well I found a formula for it but I am not really sure how to show the working for it. But here is the formula

\int \sqrt{a^2-x^2} dx =\frac{1}{2}(x\sqrt{a^2-x^2}+a^2sin^{-1}(\frac{x}{a}))+ c

I trust you that it works but we haven't been taught to use that formula yet, so I'm pretty sure he wouldn't have us use it.

I'm thinking I got something wrong in getting from the initial plug into the end simplification.

Thanks tho =P

 

[rock.freak667]

AH seems that I should actually work out the problems more...when I did it...the end result was less complicated than what you had...you did some wrong algebra in there

 

[Lanza52]

AH seems that I should actually work out the problems more...when I did it...the end result was less complicated than what you had...you did some wrong algebra in there

Hrmm...any hints? Been through it a few times and all I have changed is a positive to a negative and a negative to a positive. Haven't found anything to make it easier.

 

[rock.freak667]

Well it seems i made a mistake..twice...i keep getting what you initially had..the only help i can suggest is

\frac{\pi}{x}\int \sqrt{16-(\sqrt{3}x)^2} dx

Let (\sqrt{3})x=4sin\theta and work from there...but since it is a surface area of revolution I would expect some limits so then it would be easy from here

 

[Lanza52]

So after playing with it and using that substitution I get to
8\pi + \frac{4\pi(sin^-^1(3/2)-sin^-^1(-3/2))}{3}

Can't help but think I got something wrong in there.