JaysFan31
Jan29-07, 03:26 PM
1. The problem statement, all variables and given/known data
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)
2. Relevant 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).
3. 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?
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)
2. Relevant 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).
3. 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?