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Calculate the line integral of $ \int_{C} xy\,dx + 2y^2\,dy $, where C is composed of two parts: the arc of the circle from $ (2,0) $ to $ (0,2)$ and the line segment from $ (0,2) $to $ (0,0) $ Attempt:

For the first part (I.e circle part) let $ x = 2\cos\theta $ and $y = 2\sin\theta $ this gives $ dx = -2\sin\theta $ and $ dy = 2\cos\theta$ with $ \theta \in [0,\frac{\pi}{2}] $

Along this part of the curve C we have to compute $ \int_{C_1} (2\cos\theta)(2\sin\theta)(-2\sin\theta)\,d\theta + 2(4\sin^2\theta)(2\cos\theta)\,d\theta $, which is equal to 8/3.

Along the y axis part, I parametrized the curve in terms of t again. Obviously $x=dx=0$ and $ y= (1-t)y_1 + y_2t $ where $ y_1 = 2 $ and $ y_2 = 0 $ This reduces the line integral of C along the y axis part as $ \int_{0}^{1} 2(2-2t)(-2)\,dt $ which gives -4. Adding the two results together gives -4/3. Am I correct? Also, am I right in saying the results should be independent of parametrisation? (I.e I could have parametrized in terms of x,y etc)?

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Hmmm...I do understand your parametrization of the second path but it could be


and we take the path upwards (and then we can change the sign), so


Thus, the value of the integral is


I think that when you got into the integral you forgot to take $\,y^2\,$ , and it should be:


which is what I got above (with the sign changed, of course)

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Ok,many thanks. The choice of parameter does not matter here, correct? How could I parametrize it in terms of y (0r x)? – CAF Oct 27 '12 at 22:00
I think my parametrization is "the easiest" one as it doesn't require the geometrical-like you chose: I don't care whether it goes from (2,0) to (0,0) or the other way around as I can easily change the sign, just as I did, in case it is needed. Now, this vertical-horizontal parametrizations are pretty easy since one of the parameters is a constant (even better: zero!), so its differential $\,dx\,\,or\,\,dy\,$ vanishes. And yes: the choice of parametrization is irrelevant here, just a matter of simplicity and taste. – DonAntonio Oct 28 '12 at 2:44
Could I have integrated $ \int_{C_2} 2y^2\,dy $ by simply doing $ [\frac{2}{3}y^3] $ evaluated between 0 and 2? Side question: on a similar question to this one, I got to a stage where I had to evaluate $ \sqrt{a^6) $, a being a scalar quantity. How do I know whether to take this as $-a^3$ or $ a^3$? – CAF Oct 28 '12 at 11:09
Well, this is exactly what I did, didn't I? Of course, my parametrization allowed that! – DonAntonio Oct 28 '12 at 11:14
I was just wondering if it was sloppy notation to keep it as y? You had effectively the same thing, just in terms of t – CAF Oct 28 '12 at 11:15

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