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)?
Best Answer
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)