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Evaluate $ \int_{C} xy\,ds $ where C is the arc of the ellipse $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ in the first quadrant.

Let $x = a\cos t$ and $ y= b\sin t$ and use a substitution of $ u = a^2 \sin^2t + b^2\cos^2t $ to simply the expression under the sqrt root when dealing with $ ds$. I eventually get to $$\frac{ab}{2(a^2-b^2)} [\frac{2}{3} \sqrt{(a^2\sin^2t + b^2 \cos^2t)^3}]$$evaluated between $\pi/2$ and $0$.

Should I take $\sqrt{a^6} = a^3$ here? (my thoughts being that $ a>0 $ in first quadrant)

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You can assume that $a$ and $b$ are positive in the equation for the ellipse. (Why?) –  mrf Oct 28 '12 at 12:31
    
If a = b, then the ellipse reduces to a circle. So the radius of this circle will be a(or b) which is defined to be a positive. Ok, so in this case, I should take the root to be $ a^3 $? –  CAF Oct 28 '12 at 13:07

1 Answer 1

The correct setup is

$$\int_{0}^{2 \pi} ab\cos(t) \sin(t) \sqrt{a^2 \cos^2t+b^2\sin^2t}dt$$

As you hypothesized. Let $u = a^2 \cos^2t+b^2\sin^2t$ and $du = 2b^2\cos(t)\sin(t) - 2a^2\sin(t) \cos(t) dt$

Factor out $\cos(t) \sin(t)$, this will cancel out the same factor upstairs and in the end you should end up with

$$\int \dfrac{ab \sqrt{u} du}{2b^2 - 2a^2}$$

Factor out the constants and integrate. Then do the definite integral (which is very similar).

I leave the rest to you

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