Line integral over curve defined piecewise

CAF just a moment. 1 answers, 0 views

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

DonAntonio 10/27/2012.

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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