
1 Attachment(s)
Area of ellipse segmant
Hi,
I need help to find the area of a segmant of an elipse. In the attchment you can see an image of an ellipse with a rectangle in it. I know the parameters of the rectangle. I need area segmants I need to find area of the segmants that are not bound by the rectangle i.e. the blue bits.
Any help would be much appreciated
Dan

Since you know the area of the rectangle, just subtract it from the total area of the ellipse to find the blue section.
The area of an ellipse is given by $\displaystyle {\pi}ab$.
Where a and b are the lengths of the minor and major axes.
Say you had an ellipse $\displaystyle \frac{x^{2}}{4}+\frac{y^{2}}{16}=1$
a=2 and b=4. So, it's area would be $\displaystyle {\pi}(2)(4)=8\pi$
Of course, if we wanted to be fancy schmancy we could use integration.

Hi,
Thanks for you help. Sorry I didn't explain properly I actually need to find the area of each segmant seperatly. So I only need to work out the area of one large and one small as the other sides will be the same.
Many Thanks
Dan

Are you given any dimensions or is it a general case?.

All I have are dimensions for a rectangle, as the rectangle is been used for a set covering problem. The rectangle is just a simplification of the ellipse. So in this case the dimensions of the rectangle are 4 by 2.

Since its height is 2, the dimension of the rectangle from the origin to its top is 1.
Then, just integrate from 1 to b. Where b is the length of the minor axis of the ellipse.
Solve the ellipse equation for x and get $\displaystyle x=\frac{a}{b}\sqrt{b^{2}y^{2}}$
$\displaystyle \int_{1}^{b}\frac{a}{b}\sqrt{b^{2}y^{2}}$
This will give the area of the top segment.
The bottom one is the same.
The right and left segments can be done similarly only integrate w.r.t x and use 2 instead of 1. Integrate from 2 to a.
You could do this without calc as well.

Hi
Thanks so much for your help. I'm a little rusty on the maths front and was wondering what you meant by intergrate from 1 to b.
Many Thanks
Daniel