Evaluating Integrals on Ellipse: C and C

[teleport]

Homework Statement

(i) Evaluate

[tex]\int_C \dfrac{-ydx + xdy}{9x^2 + 16y^2} [/tex]

when C is the ellipse

[tex]\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1[/tex]

(ii) Use the ans to (i) to evaluate the integral along C' = ellipse:

[tex] \dfrac{x^2}{25} + \dfrac{y^2}{16} = 1[/tex]

Homework Equations

The Attempt at a Solution

I have done (i) but have no clue about (ii). great thnx for any help.

 

[HallsofIvy]

Okay, what did you get for (i)? I can think of a number of ways that might help you answer (ii) but since you have shown no work at all I don't know which way would be appropriate for you. Do you know Green's Theorem?

 

[teleport]

got pi/6 for i. didn't use green's for that. the other way is easier. for ii u can't use green's since it would either be too complicated or impossible to integrate. yes i know green's thm. sorry for not showing any work. i just don't know what to do for ii.

 

[HallsofIvy]

That's not at all what I get for (i). And the problem with (ii) is I don't know what theorems or methods you have available. I do notice that the integrand is defined everywhere except at (0,0) which is inside both ellipses.

 

[teleport]

checked it and still got the same. is it allowed to say 9x^2 + 16y^2 = 144 on that integral? that's something I'm using.

for ii, if u could mention some of the methods u have in mind, i might recognize it as something given in class. thnx

 

[HallsofIvy]

My apologies. I just screwed up (1) by copying part of the problem incorrectly. Yes, [itex]\pi/6[/itex] is the correct answer.

My point about the integrand not being defined at (0,0) was that you make a cut from the outer ellipse to the inner, integrate around one ellipse, then up that cut to the other and back. Then you are integrating around a curve that does NOT have (0,0) in its interior. The integral around that path will be 0, showing that the integral around the two ellipses is the same.

 

[teleport]

and why is it 0? what does the integrand not being defined at (0,0) has to do with that? by a cut do u mean to make a line connecting the 2 ellipses? could u explain more please?