0
$\begingroup$

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)?

1 Answers 1

1

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)

  • 0
    Ok,many thanks. The choice of parameter does not matter here, correct? How could I parametrize it in terms of y (0r x)?2012-10-27
  • 0
    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.2012-10-28
  • 0
    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$?2012-10-28
  • 0
    Well, this is exactly what I did, didn't I? Of course, my parametrization *allowed* that!2012-10-28
  • 0
    I was just wondering if it was sloppy notation to keep it as y? You had effectively the same thing, just in terms of t2012-10-28
  • 0
    In general, for *any* real $\,a\,$, we have $\,\sqrt{a^2}=|a|\,$ , so $\,\sqrt{a^6}=|a|^3\,$2012-10-28
  • 0
    Nothing sloppy with that as long as it doesn't confuse yourself...:)2012-10-28
  • 0
    For this question involving sqrt(a^6), the Line integral was along the ellipse $ \frac{x^2}{a^2} + \frac{y^2}{b^2} $ in the first quadrant. Does this mean I take $ \sqrt{a^6} = a^3$? Since a is positive here2012-10-28
  • 0
    I suppose so but perhaps it'd be a good idea to post a new question...after you've chosen and upvoted the above answer, if you're pleased with it.2012-10-28
  • 0
    Ok, thanks, I will do. I'll post the new question in a few moments2012-10-28
  • 0
    Thanks. Just one question: you "accepted" the above answer yet you did not *upvote* it...did you only miss this or is there any other reason?2012-10-28
  • 0
    Sorry, I am quite new to this site. What do you mean by upvote?2012-10-28
  • 0
    Where an answer begin, to the left, there are two arrows: one upwards = upvote, the other downwards=downvote. It is recommended that askers upvote answers they find useful, clear, etc., though only one answer can be "accepted" (as the best one) by the OP(=original poster)2012-10-28