1
$\begingroup$

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.

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ 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
1
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.