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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

$$x=0\,\,,\,\,dx=0\,\,\,,\,\,y=t\,\,,\,dy=dt\,\,,\,t\in[0,2]$$

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

$$\int_0^22t^2\,dt=\left.\frac{2}{3}t^3\right|_0^2=\frac{16}{3}$$

Thus, the value of the integral is

$$\frac{8}{3}-\frac{16}{3}=-\frac{8}{3}$$

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

$$\int_0^12\left[2(1-t)\right]^2(-2)dt=-16\int_0^1(1-t)^2dt=\left.\frac{16}{3}(1-t)^3\right|_0^1=-\frac{16}{3}$$

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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