# Integral of an Elipse

• Aug 14th 2010, 07:35 PM
leomylonas
Integral of an Elipse
Hi... again :p

I need to calculate the VOLUME of an elipse, but I have no idea how to do it... I assume it would require integration though.

The cross-sectional function for my elipse is:

$\displaystyle \sqrt{((1-(x-4)^2)/16)*9} ; 0<=x<=6$

and

$\displaystyle -\sqrt{((1-(x-4)^2)/16)*9} ; 0<=x<=6$

which gives:

Attachment 18577

So as you can see, I'm trying to calculate the volume from 0 to 6 of this elipse.

I thought about calculating the volume of the full elipse, then find the proportion of volume i need, but that wouldnt work because it isnt a uniform shape... woudl it?

• Aug 14th 2010, 09:49 PM
Failure
Quote:

Originally Posted by leomylonas
Hi... again :p

I need to calculate the VOLUME of an elipse,

I suppose you mean the AREA.

Quote:

but I have no idea how to do it... I assume it would require integration though.
Not really, in this particular case. You can get by with just a little geometry+trigonometry, since you can determine the area of that part of the ellipse by multiplying the corresponding area of the circle over its major axis (with diameter 8) by the factor 1/2 (which is the ratio between minor and major axis of the ellipse).
• Aug 14th 2010, 11:04 PM
leomylonas
Quote:

Originally Posted by Failure
I suppose you mean the AREA.

Nope. I mean volume. After a bit more research, I found out what I want to do is called "Volume of Revolution"

Volume of A Solid of Revolution

I think I have done this right now.

http://www.mathhelpforum.com/math-he...1&d=1281854896

However now i have one that is a bit more difficult.

$\displaystyle f(x) = 1/(0.1(x-8)) +0.5; -1.5<=x<=6$

Which means to find the volume i have to calculate this:

$\displaystyle \int_{-1.5}^{6} \pi*(1/0.1(x-8) +0.5)^2 dx$

I've got no idea how to easily integrate that. I could expand it out, but that makes it even harder. Does anyone wanna have a crack at working that one out?

(Oh, and with the MATH tags, how do you can I make a fraction display properly?)

Thanks
• Aug 15th 2010, 01:18 AM
HallsofIvy
You also mean "ellipsoid", not "ellipse".

To get a fraction, put { and } around what ever you want in the numerator and what ever you want in the denominator. \frac{x^2+ 3xy}{2x- y} gives $\displaystyle \frac{x^2+ 3xy}{2x- y}$.
• Aug 15th 2010, 03:29 AM
mr fantastic
Quote:

Originally Posted by HallsofIvy
You also mean "ellipsoid", not "ellipse".

To get a fraction, put { and } around what ever you want in the numerator and what ever you want in the denominator. \frac{x^2+ 3xy}{2x- y} gives $\displaystyle \frac{x^2+ 3xy}{2x- y}$.

Ah, to think that I have lived to see the day that HoI would give help on posting with latex (off topic I know but I just had to say it (Wink) )
• Aug 15th 2010, 03:44 AM
Unknown008
I don't know... but what if you tried to move a 2d ellipse so that the centre of the ellipse is at the origin?

Then, you integrate, using the theory of volume of rotation:

$\displaystyle Volume = \pi \int^a_{-a} y^2 dx$

where a and -a are the points where the ellipse cut the x-axis and the curve y is always positive...
y is then another curve with another formula.

Or is it not possible this way?