1. ## Line integral

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)

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).
(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?

2. 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)?

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)

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).
(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?
I am assuming it moves along the ellipse only.

And I am also assuming it is positively oriented (abusing terminology).

The the part can be described as $x=a\cos t$ and $y=b\sin t$ for $0\leq t\leq \frac{\pi}{2}$. The reason why it is cosine for "x" because it starts from the right and goes to the left as I am assuming.