Lanza52
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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?
(\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?