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.