Line Integral

  • Thread starter JaysFan31
  • Start date
  • #1
JaysFan31

Homework Statement


This is my problem:
Compute the following three line integrals directly around the boundary C of the part R of the interior ellipse (x^2/a^2)+(y^2/b^2)=1 where a>0 and b>0 that lies in the first quadrant:
(a) integral(xdy-ydx)
(b) integral((x^2)dy)
(c) integral((y^2)dx)


Homework Equations


I used parametrisation (x=acost and y=bsint) for the arc of the ellipse.
C is the curve r=(acost)i + (bsint)j (0 less than or equal to t less than or equal to pi).


The Attempt at a Solution


(a) integral(xdy-ydx)=integral from 0 to pi((acost)(bcost)dt)- integral from 0 to pi((bsint)(-asint)dt)=(ab)pi/2-(-ab)(pi)/2=ab(pi)

(b) integral((x^2)dy)=integral from 0 to pi((acost)(acost)(bcost)dt)=0.

(c) integral((y^2)dx)=integral from 0 to pi((bsint)(bsint)(-asint)dt)=-(4/3)a(b^2)

Could anyone check these and see if they are right?
 

Answers and Replies

  • #2
JaysFan31
Actually, should all the bounds be from 0 to pi/2 instead of 0 to pi (since I am looking at only the first quadrant)?
 

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