I have a line integral of kind 2. I want to use Green's theorem to solve it. I am not sure if I am setting it right so I want to ask for help to verify if I set up the integral correct. So the integral is given as follows : $$\iint (2x)dx+3(yx)dy$$ with parameters : $$C:x = 4\cos(2t) \ , \ y=3\sin(2t)$$ I found the region and its an ellipse: $$\frac{x^2}{4}+\frac{y^2}{3}=1$$ And I use polar coordinates parameters : $$x = 4r\cos 2t\\ y = 3r\sin 2t\\ dy\ dx = 12r \ dr\ dt$$ So I apply the greens theorem $$\int_C P(x, y)dx + Q(x, y)dy = \iint_S \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dxdy$$ and I get $$\iint_S 3y dydx$$ $$\int_0^{2 \pi} \int_0^1(9r\sin(2t))(12r)drd \theta $$ Is this the right way to evaluate this integral with the Green's theorem ?
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1Can you Please check your question once more? – Kumar Jun 19 '19 at 06:17
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@Kumar Whats wrong with the question ? If you tell me ill fix it. Thanks. – Boris Borovski Jun 19 '19 at 06:23
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I think it should be $\iint (2x+3yx)dxdy$ – Kumar Jun 19 '19 at 06:25
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@Kumar After applying the Green's theorem we swap the derivatives and $2x$ is in terms of $dy$ then 2x is constant in terms of dy and you get the new integral just with $3y$. In other words our $P$ is $2x$ and $Q$ is $3y$ before we apply the formula. Maybe i should clear this out with one more line there. – Boris Borovski Jun 19 '19 at 06:35
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Can you tell me how you got $dx dy = 12 rdr dt$ ? – xrfxlp Jun 19 '19 at 10:49
1 Answers
There are three things you have done here wrong, they are:
Equation of the ellipse would be $$ \cfrac{x^2}{16} + \cfrac{y^2}{9} = 1$$ and it can be verified by substituting $x = 4 \cos(2t), y = 3 \sin(2t)$ in the equation.
As you taken $x = 4r cos(2t), y = 3r sin(2t)$. Since$$ dx = \cfrac{\partial x}{\partial r} dr + \cfrac{\partial x}{\partial t}dt $$ and same for $dy$, $$\begin{align} &\Rightarrow dx = 4 \cos(2t) dr - 8r \sin(2t) dt \\ & \Rightarrow dy = 3 \sin(2t) dr + 6r \cos(2t) dt \end{align} $$ whose exterior product is $$dx dy = 24 r dr dt$$
Limit of $t$ would be from $ 0$ to $ \pi$ , you can check that by placing $\pi$ in $x = 4r \cos(2t) $, it reverts back to $4r$ i.e. it has completed one round.
In the last integral, $d \theta$ is not defined.
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@jay Mishra Thank you i can't understand this formula how dxdy is calculated. For the ellipse i just forgot the squares. – Boris Borovski Jun 19 '19 at 14:01
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Btw the function under the integral is this right just $dxdy$ part is wrong ? Yes it should be can't. – Boris Borovski Jun 19 '19 at 14:12
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https://math.stackexchange.com/questions/1636021/rigorous-proof-that-dx-dy-r-dr-d-theta See this. – xrfxlp Jun 19 '19 at 14:15
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I mean this integral $\int_0^{2 \pi} \int_0^1(9r\sin(2t))(12r)drd \theta$ , from where is $ \theta$ ? – xrfxlp Jun 19 '19 at 14:15
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i used the theorem before going to polar coordinates and $2x$ is differentiate with respect to $dy$ so its constant and i left only with 3y and i substitute it. – Boris Borovski Jun 19 '19 at 14:20
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Thank you a lot for the link i got it i need to calculate the determinant for the jacobian and i get the $24rdr d\theta$ . – Boris Borovski Jun 19 '19 at 14:37
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You can visualize that just as area of infinitesimal parallelogram. Anyway, cool. – xrfxlp Jun 19 '19 at 14:38
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One last very important for me question if i am not bothering you why when we have ellipse that is centered at 0 we don't integrate over $2 \pi$ and there are no constraints for $cost$ and $sint$ – Boris Borovski Jun 19 '19 at 14:41
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