Consider the ellipse:(adsbygoogle = window.adsbygoogle || []).push({});

[tex] (\frac{x}{2})^2 + y^2 = 1 [/tex]

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.

[tex] y = \sqrt{1-\frac{x^2}{4}} [/tex]

Then find the derivative.

[tex]y' = \frac{-x}{2\sqrt{4-x^2}} [/tex]

Then plug into the formula for surface of revolution.

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

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

Plenty of simplifications later yields

[tex] \frac{\pi}{2} \int \sqrt{16-3x^2} dx [/tex]

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 [tex]\int 4x dx [/tex] or something simple like that. So that makes me think I'm wrong.

Any help?

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# Homework Help: Surfaces of Revolution

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