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Use Green’s Theorem to evaluate the line integral along the given positively oriented curve.

$$\int_c y^3 \, dx - x^3 \, dy, C \text{ is the circle } x^2+y^2=4$$

Ok, so I'm not sure how to approach this problem. I can easily find $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$, but I'm not sure which approach to take after that.

Am I supposed to parameterize by making $y = 2\sin t$ and $x = 2\cos t$ ? and then do I replace dxdy with $-4\sin t\cos t\, dt$? and when I plug those in does it become a single integral from $0$ to $2\pi$? And how would I integrate that? It seems like this would leave me with an over-complicated integral. Would it be faster to switch it to polar coordinates? if so how? I'm not really sure if this problem is implying I use a certain method and I'm not sure if I'm doing that method correctly. I looked in the book for similar examples but I think they vary slightly so I'm not sure which approach I'm supposed to take.

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    $\begingroup$ When applying Green’s theorem, you should be integrating over the entire disk, not it's boundary. $\endgroup$ – amd Dec 5 '17 at 5:16
  • $\begingroup$ so am I supposed to switch it to polar coordinates? $\endgroup$ – 2316354654 Dec 5 '17 at 5:17
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Green's theorem tell us that $$\int_C y^3 \, dx- x^3\, dy = \iint_{x^2+y^2 \leq 4} \left(\frac{\partial (-x^3)}{\partial x}\right)- \left(\frac{\partial y^3}{\partial y}\right) \,dx \, dy$$

While it is not a must to use polar coordinate, I strongly Iencourage you to do so as the form becomes elegant once you use polar coordinate.

Note that $\,dx\,dy =r \, dr \, d\theta $ when you change to polar coordiante.

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  • $\begingroup$ goodness gracious, I wrote down the integral from problem 8 while using the circle C from problem 9 this whole time and was wondering why it came out so weird. >_< I'm losing it $\endgroup$ – 2316354654 Dec 5 '17 at 5:48
  • $\begingroup$ i keep getting -96 pi but its -24 pi.. can't tell what I'm doing wrong. I reduced it to the integral $3\int\int((-1)(x^2+y^2))rdrd\theta$ and the outer integral is 0 to 2pi and inner is 0 to 4.. what am I doing wrong? $\endgroup$ – 2316354654 Dec 5 '17 at 6:05
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    $\begingroup$ The radius is $2$ rather than $4$. $\endgroup$ – Siong Thye Goh Dec 5 '17 at 6:07
  • $\begingroup$ oh yes... do I replace $x^2 + y^2$ with 2 also though? because if I leave that as 4 I still get -48pi $\endgroup$ – 2316354654 Dec 5 '17 at 6:10
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    $\begingroup$ I should change my previous statement, the radius for the region interest is from $r=0$ to $r=2$. We can't replace $x^2+y^2$ with $2$, we replace $x^2+y^2$ with $r^2$. $\endgroup$ – Siong Thye Goh Dec 5 '17 at 6:12

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