- #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?