Compute the scalar line integral for () = on , where is oriented counterclockwise and comprises : the ellipse $\ \frac{^}{ }+ \frac{^}{}= $ in the first quadrant, : the line segment from $\ (, )$ to $\ (, )$
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Welcome to MSE! Please tell us: What have you tried? Rather than post a question and wait for solutions, please help us help you. Show your working, so someone can help troubleshoot your problem. That makes this site so much more than a glorified calculator. – MathAdam Apr 29 '17 at 17:30
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I'm completely confused on how to do the first part, but the second part I tried parametrization and evaluated the integral. – Mo Bhuiyan Apr 29 '17 at 17:40
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The secret to all these types of questions is to parametrise your function $f(x)$ and path we need to integrate over. You say in a comment that you have done the second part so, for the first part the ellipse:
Note that we parametrise $\frac{x^2}4 + \frac{y^2}9 = 1$ in the first quadrant by $x=2\cos(\theta) , y = 3\sin(\theta)$ with $\theta \in [0, \frac{\pi}2] $ -convince yourself that this parametrises the desired part of the ellipse!! Then the question reduces to integrating $f(x) = xy = 6\cos(\theta)\sin(\theta)$:
$\int_0^{\frac{\pi}2} 6\cos(\theta)\sin(\theta) d\theta$
This integral is now done with a simple application of the sine double angle formula, hope that helps!!
Vincenzo Tibullo
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SEWillB
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Do you mind telling me the final answer? I just want to double check. – Mo Bhuiyan Apr 29 '17 at 18:11
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$\int_0^{\frac{\pi}2} 6\cos(\theta)\sin(\theta) d\theta = 3\int_0^{\frac{\pi}2} \sin(2\theta) d\theta = \frac32[-\cos(2\theta)]_0^{\frac{\pi}2} = \frac32((1)-(-1))=3 $ – SEWillB Apr 29 '17 at 18:24
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