Area of A Sector of An Ellipse

Date: 02/28/98 at 08:29:29
From: Phill Brady
Subject: Area of a sector of an ellipse

I need a formula for the area of an ellipse sector. I have the 
semiminor and major axes and the angles of the 2 vectors bounding the 
sector. How do I work out the area?

Date: 02/28/98 at 16:13:08
From: Doctor Sam
Subject: Re: Area of a sector of an ellipse

Phill,

You can find the area of a sector of an ellipse by representing the 
ellipse in polar coordinates and then integrating.  

If the ellipse has semi-major axes of A and B in the x- and y-
directions, then the ellipse can be represented as 

          x = AcosT  and 
          y = BsinT.

Since

   x^2 + y^2 = r^2

we can get  

   r^2 = A^2 cos^2 T + B^2 sin^2 T.

Now, area in polar coordinates is given by an integral:

   (1/2) INT(r^2 dT) from T1 to T2

T1 and T2 are the angles of the vectors defining the sectors that you 
mention. So you need to evaluate 

   area = (1/2) INT(A^2 cos^2 T + B^2 sin^2 T dT) from T1 to T2
    
        = (1/2) INT(A^2 cos^2 T dT) from T1 to T2

             + (1/2) INT(B^2 sin^2 T dT) from T1 to T2

        = (1/4) A^2[ T2 - T1 + sinT2 cosT2 - sinT1 cosT1 ]

             + (1/4) B^2[ T2 - T1 - sinT2 cosT2 + sinT1 cosT1 ]

This can be simplified a little by regrouping:

   area = (T2-T1)(A^2+B^2)/4 + (sinT2 cosT2 - sinT1 cosT1)(A^2-B^2)/4 

I hope that helps.

-Doctor Sam, The Math Forum
Check out our web site http://mathforum.org/dr.math/   

Date: 03/27/2000 at 11:57:24
From: Hume Peabody
Subject: Area of ellipse sector

I need to find the area of a sector of an ellipse between theta1 and 
theta2. The solution posted is incorrect. This can be shown by the 
simple fact that a full ellipse has an area of PI*A*B. The posted 
solution is:

     area = (T2-T1)(A^2+B^2)/4 + (sinT2 cosT2 - sinT1 cosT1)(A^2-B^2)/4

For T1 = 0 and T2 = 2*PI, this yields (A^2 + B^2)*PI/2

I can work the radius down as a function of theta, but cannot find a 
closed form solution from 0 to 2*PI for the integration.

     R = A*B/SQRT(B^2*cos^2(Theta) + A^2*sin^2(Theta))

I can integrate in rectangular coordinates and also find the area of 
the remaining triangle to get the sector area, but I was looking for a 
more elegant solution.

Thanks for any help.

Hume

Date: 03/28/2000 at 12:20:17
From: Doctor Peterson
Subject: Re: Area of ellipse sector

Hello, Hume.

Thanks for letting us know about the error in the archives. It looks 
like Dr. Sam forgot that the variable T in the parametric form is not 
the angle, and therefore integrated with respect to the wrong variable.

Replacing dT with

               A sec^2(T)
     dTheta = ------------
              B + A tan(T)

would give the correct integral; or instead we can use your integral in 
terms of Theta and integrate it by substituting u = tan(Theta). 
Eventually we get

     Area[0,Theta] = (AB/2) arctan(A/B tan(Theta))

so that

     Area[Theta1,Theta2] = (AB/2)[arctan(A/B tan(Theta2) -
                                   arctan(A/B tan(Theta1)]

(In evaluating this, you have to choose the right branch of the arctan; 
be careful!)

This can also be found without integration very easily. If we transform 
the ellipse to a circle of radius B by "compressing" in the X direction

           *************               *****
       ****      |     /****        ***  | /***
     **          |  /Theta  **    **     |/T   *
    *------------+------------*   *------+------*
     **          |          **    **     |     *
       ****      |      ****        ***  |  ***
           *************               *****

It turns out that the angle in the circle is exactly the T in the 
parametric equation, which is

     T = arctan(A/B tan(Theta))

The area of the sector between the x-axis and the angle Theta is b/a 
times the area of the sector between x-axis and angle T in the circle, 
which gives our formula.

You can decide which is most elegant.

For some related formulas, though not the area of a sector, see this 
page in World of Mathematics:

   http://mathworld.wolfram.com/Ellipse.html   

We'll be sure to correct the archives.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/