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Question: Test Green's Formula in the case where $M = x^2 - y^2$ and $N=2xy$ and $\Omega$ is the triangle with vertices $(0, 0), (1,1), (2, 0)$.

My attempt: $$\iint_\Omega \Big(\dfrac{\partial M}{\partial x} - \dfrac{\partial N}{\partial y} \Big) \,dx\,dy = \iint_\Omega (2x - 2x) \,dx\,dy = 0,$$ So must the following integral be zero as well: $$\int_\Gamma (x^2-y^2) \,dy + 2xy \,dx = 0+ \int_{\frac13}^{\frac23} (-4+12t)(-3) \,dt + 0 + \int_{0}^{\frac13} 2(9t^2)(3) \,dt + \int_{\frac13}^{\frac23} 3(2)(3t)(2-3t) \,dt +0= \dfrac{84}{9} \ne 0 \ !!$$ Where the first three integral is for ... dy and the last three is for ... dx. Where am I doing wrong?

I used $(3t,3t)$ for from $(0,0)$ to $(1,1)$; $(3t,2-3t)$ for from $(1,1)$ to $(2,0)$; and, $(6-6t,0)$ for from $(2,0)$ to $(0,0)$.

By the book the Green's Formula is $$\int_\Gamma (Mdy+Ndx)=\iint_{\Omega}\left(\frac{\partial M}{\partial x}-\frac{\partial N}{\partial y}\right)dx\,dy.$$

I calculated the line integrals clockwise but the difference is a minus in final result; so no big deal!

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  • $\begingroup$ I think you messed up $M$ and $N$. I am guessing that you should be integrating $Mdx+Ndy$ over $\Gamma$, which is the same as integrating $\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}$ over $\Omega$. $\endgroup$ – user593746 Oct 7 '18 at 15:59
  • $\begingroup$ @Zvi, yes I edited. $\endgroup$ – user231343 Oct 7 '18 at 16:00
  • $\begingroup$ Shouldn't the first integral be $\int_{1/3}^{2/3}6(2-9t^2)dt$? $\endgroup$ – user593746 Oct 7 '18 at 16:06
  • $\begingroup$ Shouldn't the last integral be just $0$? $\endgroup$ – user593746 Oct 7 '18 at 16:07
  • $\begingroup$ If you fixed these two integrals, you should have that the line integral equals $-\frac23+\frac23+0=0$. $\endgroup$ – user593746 Oct 7 '18 at 16:08
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You can parametrize the boundary of your triangle by $\gamma_1(t) =(2t,0)$, $\gamma_2(t)=(2-t,t)$ and $\gamma_3(t) = (1-t,1-t)$ over $[0,1]$. Then the pullback of $\omega$ under these three functions are

$$\omega_1 = 0,\quad \omega_2 = 2(t^2-4t+2)dt,\quad \omega_3= -2(t-1)^2 dt.$$

But note that $\omega_2 = 2((t-1)^2 - (2t-1))$ so that the total pullback is the linear function $\omega_1+\omega_2+\omega_3 = -2(2t-1)$ which integrates to $0$ over $[0,1]$.

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