Expert: Paul Klarreich - 2/22/2006

Question
Hi, im a little stuck with this question:

Q: Use a line integral round the ellipse (x^2/a^2)+(y^2/b^2) = 1 to find the area of the region
that it encloses.

I think that Green's theorem is needed, or a result from Green's theorem, but i dont know how to apply this. i'm pretty sure the answer is pi*a*b tho! any help would be greatly appreciated!

Answer
James Waller Asks in Category Calculus:
Subject:  greens theorem
Question:  Hi, im a little stuck with this question:

Q: Use a line integral round the ellipse (x^2/a^2)+(y^2/b^2) = 1 to find the area of the region that it encloses.

I think that Green's theorem is needed, or a result from Green's theorem, but i dont know how to apply this. i'm pretty sure the answer is pi*a*b tho! any help would be greatly appreciated!
----------------------------------------------

Yes, you can use GT.  Basically it says that an integral of a function over an area is the same as the line integral of (some related) function about the boundary.  If all you want is the area itself, then the function to be integrated is f(x,y) = 1.  (Wow, that's deep!)

GT says:

{{
|| (dQ/dx - dP/dy) dA =
}}

{
| P dx + Q dy
}

Since in the first one, the area integral, you want to integrate 1,  (Yes, one.) all you have to do is select P and Q so that:

dQ/dx - dP/dy = 1

Of course there are lots of ways to do this, but one easy way is to select:

Q = x,  P = 0.

WARNING: THE MATERIAL BELOW MAY CONTAIN FRACTIONS AND OTHER MATERIAL INAPPROPRIATE FOR CERTAIN COMPUTING SYSTEMS.  BE SURE TO VIEW IT IN A FIXED-SIZE FONT, SUCH AS COURIER.

Now you just take your line integral as:

{               {  
| P dx + Q dy = | 0 dx + x dy
}               }

Now on the ellipse, you use parametric equations to write x and y:

x = a cos t,  
y = b sin t  which gives  dy = b cos t dt

The integral is:  (evaluated between 0 and 2 pi)

{                        {  
| a cos t b cos t dt = ab| cos^2 t dt
}                        }

You do cos^2 t using the old 'half-angle' trick.
         1 + cos(2t)  
cos^2 t = ------------
              2

 {  1 + cos(2t)
ab| ------------- dt =
 }       2

   t    cos 2t
ab(--- + ------)
   2      4

Now when you evaluate between t=0 and t=2 pi, the cosine disappears, and all you have is:

ab(2pi/2) = pi ab, as you suspected.