I'm trying to find the area of an ellipsoid from x=-a to x=a using y=b(1-x^2a^-2)^1/2. I know the formula and plugged in the function, but I don't know how to get the anti derivative from this. pi~-a to a of [b(1-x^2a^-2)^1/2]^2dx. Could someone please help me with this antiderivative?
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Area of an ellipsoid
- Thread starter twinmom
- Start date
An ellipsoid is 3-dimensional.(kind of like an egg, for lack of a better description). Are you sure you don't mean an ellipse?. You would find the volume of an ellipsoid and the area of an ellipse.
The equation of an ellipsoid is given by :
\(\displaystyle \L\\\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1\)
An ellipse is:
\(\displaystyle \L\\\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
The equation of an ellipsoid is given by :
\(\displaystyle \L\\\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1\)
An ellipse is:
\(\displaystyle \L\\\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
Hello, twinmom!
Could you clarify?
You said "area", but \(\displaystyle \pi\int y^2\,dx\) gives us volume.
I'll assume you want the volume of an ellipsoid of revolution.
Did you square it?
You get: \(\displaystyle \L\:\pi\int^{\;\;\;a}_{-a}b^2\left(1\,-\,\frac{x^2}{a^2}\right)\,dx \;= \;\pi b^2\int^{\;\;\;a}_{-a}\left( 1\,-\,\frac{1}{a^2}x^2\right)\,dx\)
. . And you can't integrate that?
Could you clarify?
You said "area", but \(\displaystyle \pi\int y^2\,dx\) gives us volume.
I'll assume you want the volume of an ellipsoid of revolution.
I'm trying to find the volume of an ellipsoid from \(\displaystyle x\,=\,-a\) to \(\displaystyle x\,=\,a\)
. . using the semiellipse: \(\displaystyle \,y\:=\:b\left(1\,-\,\frac{x^2}{a^2}\right)^{\frac{1}{2}}\)
I don't know how to get the anti derivative from this:
. . \(\displaystyle \L\pi\int^{\;\;\;a}_{-a}\left[b\left(1\,-\,\frac{x^2}{a^2}\right)^{\frac{1}{2}}\right]^2\,dx\) . . . you don't?
Did you square it?
You get: \(\displaystyle \L\:\pi\int^{\;\;\;a}_{-a}b^2\left(1\,-\,\frac{x^2}{a^2}\right)\,dx \;= \;\pi b^2\int^{\;\;\;a}_{-a}\left( 1\,-\,\frac{1}{a^2}x^2\right)\,dx\)
. . And you can't integrate that?
I'm sorry, yes I meant to say volume, not area. Regardless, my skills at integrations are not very sharp. I tried to integrate what you had in your response and I got: pib^2 from -a to a x-1/3x^3(-1/a). I know that can't be right; what am I doing wrong when integrating x^2/a^2?