My Math Forum In dire need of some help on a Green's theorem problem

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 July 2nd, 2018, 05:43 PM #1 Newbie   Joined: Jul 2018 From: Vancouver, BC, Canada Posts: 1 Thanks: 0 In dire need of some help on a Green's theorem problem Denote by C the frontier of the triangle with vertices (0,0), (1,0) and (0,1). Show that the line integral of x⁴dx + (xy) dy = 1/6. I know that I'm supposed to apply the Green's theorem, but I'm having a somewhat hard time at setting up the curve C. Any help would be appreciated.
 July 2nd, 2018, 06:22 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,202 Thanks: 1157 look at example 1 here
 July 3rd, 2018, 04:41 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 You are having trouble "setting up the path"? It's just three line segments, the line x= 0 from y= 0 to y= 1, the line y= 0 from x= 0 to x= 1 and the line y= x from (0, 0) to (1, 1)! Or do you mean you are having trouble finding parametric equations for the path? The problem does not ask you to do that! Green's theorem says that $\displaystyle \oint L dx+ Mdy= \int\int \left(\frac{\partial M}{\partial x}- \frac{\partial L}{\partial y}\right) dxdy$. That is, that the integral of $\displaystyle L dx+ Mdy= x^4 dx+ xy dy$ around a closed path, here the triangle with vertices at (0, 0), (1, 0), and (0, 1), is the same as the integral of $\displaystyle \frac{\partial M}{\partial x}- \frac{\partial L}{\partial y}= y- 0= y$. That is $\displaystyle \int_{y= 0}^{y= 1}\int_{x= 0}^{1- y} y dxdy=$$\displaystyle \int_0^1 y(1- y)dy= \int_0^1 y- y^2 dy=$$\displaystyle \left[\frac{y^2}{2}- \frac{y^3}{3}\right]_0^1= \frac{1}{2}- \frac{1}{3}$.
 July 3rd, 2018, 04:55 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 The problem asks you to use Green's theorem to find the path integral - that is, integrate over the interior of the triangle. It does NOT ask you to verify that Green's theorem is correct here - that is, that the path integral is the same as the integral over the interior. But had it asked that, it certainly is not difficult to do that path integral. Part of the path is the x-axis, y= 0, from x= 0 to x= 1. We can take x itself as parameter: x= t and y= 0 so that dx= dt and dy= 0. $\displaystyle L dx+ Mdy= x^4dx+ xydy= t^4 dt$. The integral on that part of the triangle is $\displaystyle \int_0^1 t^4 dt= \left[\frac{1}{5} t^5 \right]_0^1= \frac{1}{5}$. Another part of the path is the line y= 1- x, from x= 1 to x= 0. We can again take x as parameter: x= t and y= 1- t so dx= dt, dy= -dt. The integral becomes $\displaystyle \int_1^0 t^4dt+ (t- t^2) dt= \int_0^1 (t^2- t- t^4)dt= \left[\frac{t^3}{3}- \frac{t^2}{2}- \frac{t^5}{5}\right]_0^1= \frac{1}{3}- \frac{1}{2}- \frac{1}{5}= \frac{10- 15- 6}{30}= -\frac{11}{30}$. Finally, the last part of the path is the y-axis, x= 0, from y= 1 to y= 0. Now take y as parameter: x= 0, y= t, so dx= 0, dy= t. The integral becomes $\displaystyle \int_1^0 0^4 dt- t(0)dt= \int_0^1 0 dt= 0$. The integral around the path is $\displaystyle \frac{11}{30}- \frac{1}{5}= \frac{11- 6}{30}= \frac{5}{30}= \frac{1}{6}$. But it is far easier to use Green's theorem to find that path integral by integrating over the interior of the triangle. Last edited by skipjack; July 3rd, 2018 at 10:29 AM.

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