- #1
Lanza52
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Consider the ellipse:
[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?
[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?