Question

Green's Theorem: Limits for r in polar coordinates?

Answer 1

\[\int\limits_{}^{} xy^2dy-x^2ydx\]
\[x^2 + y^2 = r^2\]
Does this means that when I convert it into polar coordinates the limit for r should be between 0 and 1? My solution for this problem was \[\pi/2\]

Answer 2

\[M(x,y)=-x^2y\]\[N(x,y)=xy^2\]\[\frac{\partial M}{\partial y} = -x^2\]\[\frac{\partial N}{\partial x} = y^2\]\[0 \leq \phi \leq 2\pi\]\[0 \leq r \leq 1\]\[x=rcos(\phi)\]\[y=rsin(\phi)\]\[\int\limits_{}^{} \int\limits_{}^{} (\frac{\partial N}{partialx}- \frac{\partial M}{\partial y}) dA\]\[\int\limits_{}^{}\int\limits_{}^{} (y^2-x^2)dA\]\[\int\limits_{2\pi}^{0}\int\limits_{1}^{0} (r^2\sin^2(\phi)-r^2\cos^2(\phi)) r dr d\phi\] IS this how it should go?

Answer 3

Somehow I switched the limits of integration up there. Sorry.

Answer 4

Here's what I got. But I don't know where the limits of the integration come from.

Answer 5

Looks like what you have, except you lost a minus sign

Answer 6

phi that's exactly like mine solution ----> \[\frac{\pi}{2}\] What about the minus sign?

Answer 7

your M starts with -, but when you plug it into the integral you dropped it.

Answer 8

nvm, looks like a typo, because you got pi/2

Answer 9

Ok, anyway thank you! You helped me cause I was uncertain. I'll click "Good answer" now.

Answer 10

But where do the limits come from?

Answer 11

Limits of integration? Well what was worrying me is can I take \[0 \leq r \leq 1\]

Answer 12

I now see what you mean by minus sign. That's my mistake. Ugh :D

Answer 13

Let R be a region in the xy-plane that is bounded by a closed, piecewise smooth curve C.

so what is R for your problem. Apparently region inside the unit circle?

Answer 14

yes it's a circle \[x^2 + y^2 = r^2\]

Answer 15

Is there more to the question? Did they give any background on what contour they are integrating around?

Answer 16

Besides \[x^2 + y^2 =r^2\] , no. The problem was just to solve it using Green's Theorem.

Answer 17

In that case, I would make the limits of r 0 to R_0

Answer 18

R_0?

Answer 19

\( R_0 \) the radius of the circle you are integrating over. You can't say it is 1. Of course, I may be missing something....

Answer 20

Well that's what was worrying me too. since we don't know how "big" is r the 1 in the limit was worrysome. So, I'll use R_0 instead it's more accurate.

Answer 21

If you ever get the answer, post it here. I'm curious.

Answer 22

Sure. Thanks again! I'm off to bed now. :)