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$F(x,y) = (y^3-6y)i + (6x-x^3)j$

a. Using Green's Theorem, find the simple closed curve C for which the integral $ ∳F \cdot dr $ (with positive orientation) will have the largest positive value.

b. Compute this largest possible value.

I'm quite certain that this is just $ \iint Nx-My $ $dA$ but I do not know how to find the bounds in this scenario for both integrals. Though I'm also sure that this problem can also be done using just $ ∳ F \cdot dr $ as there is an equation given for F.

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You are correct that if we integrate around a closed curve $C$ that bounds a region $\Omega$, then $$ \oint F \cdot dr = \iint_{\Omega} (N_x - M_y) dA. $$ Here, we have $$ (N_x - M_y)(x,y) = (6 - 3x^2) + (6-3y^2) = 12 - 3(x^2 + y^2). \tag{1} $$ In order to maximize $\iint_{\Omega} (N_x - M_y) dA$, we want $\Omega$ to include all the points where $ N_x - M_y$ is positive and none of the points where it is negative. From (1) it is apparent that $(N_x - M_y)(x,y) \geq 0$ if and only if $x^2 + y^2 \leq 4$, i.e. $(N_x - M_y)(x,y) \geq 0$ if and only if $(x,y)$ is in a disk of radius $2$ centered at the origin.

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  • $\begingroup$ So that would mean that the bounds would be from 0 to 2 for both integrals, correct? $\endgroup$ – user63266 Jan 31 at 3:07
  • $\begingroup$ No, if you wanted to do it in Cartesian coordinates, the bounds would be from $-2$ to $2$ for $x$ and $ \pm \sqrt{4-x^2}$ for $y$ (or the same but with the roles of $x$ and $y$ reversed). I would not use Cartesian coordinates for part b though. It is much better suited for polar coordinates. $\endgroup$ – Jordan Green Jan 31 at 3:10
  • $\begingroup$ Ok, I suspected as much, so θ would be from 0 to 4π, and then r would be 0 to 2? Or would it be from -4π to 4π and -2 to 2 for θ and r respectively? $\endgroup$ – user63266 Jan 31 at 3:24
  • $\begingroup$ You’re right that $r$ should go from $0$ to $2$, but the limits for theta should be taken to be $0$ and $2 \pi$ (or, alternatively, any real numbers with a difference of $2 \pi$). $\endgroup$ – Jordan Green Jan 31 at 3:27
  • $\begingroup$ Ok, I thought that because the disk was radius 2, that theta should be from 0 to 4π, but that makes sense because it's only one rotation around the unit disk with radius 2. Thank you so much for the help! $\endgroup$ – user63266 Jan 31 at 3:29

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