# Correctly setting up flux integrals

My question has to do with picking the correct limits for integration. I thought I had it figured out well, but I had an interesting issue with a homework problem. The problems were about Green's Theorem. As you know the following is the theorem for Flux integrals $\int \int_R \frac{\partial M}{\partial x} + \frac{\partial N}{\partial y} dx dy$. This is a homework problem which I worked but came to the wrong answer. The path $C$ along the curve $y=x^2, (0,0) \rightarrow (1,1)$ and then $x=y^2, (1,1) \rightarrow (0,0)$. Because the picture was drawn this way, I set up my integral like this

$$\begin{array}{rcl} \mathbf{F} & = & \langle xy+y^2, x-y \rangle \\ \end{array} \\ \begin{array}{cc} M = xy+y^2 & \frac{\partial M}{\partial x} = y \\ N = x - y & \frac{\partial N}{\partial y} = -1 \\ \end{array} \\ \int_{1}^{0}\int_{y^2}^{\sqrt{y}} \frac{\partial M}{\partial x} + \frac{\partial N}{\partial y} dxdy = \int_{1}^{0}\int_{y^2}^{\sqrt{y}} y-1 dxdy$$

This integral resulted in the correct answer by magnitude but the wrong sign. I found that by using this integral

$$\int_{0}^{1}\int_{y^2}^{\sqrt{y}} y-1 dxdy$$

Yeilded the correct answer. I used the limits I did originally because I was picturing "a particle" or something similar moving along the path in the path indicated. However, I got the wrong answer because of my choice of limits. Why did I get the limits incorrect, or did I? As I write this question, and think through what I've learned from past courses, should I have negated the integral? Like this?

$$- \int_{1}^{0}\int_{y^2}^{\sqrt{y}} y-1 dxdy$$

Thanks,
Andy

• When you set up the double-integral, make sure to have the limits in the correct order. The region can be described by the inequalities $0\leq y \leq 1$ and $y^2 \leq x \leq \sqrt{y}$, so the 0 is a lower limit and the 1 is an upper limit for the $y$ integral. – user197427 Apr 20 '15 at 18:08
• I think I see it. It seems I may have been confusing the limits by the picture provided (the book shows a picture of the region with arrows pointing from $(0,0) \rightarrow (1,1)$ for $y=x^2$ and $(1,1) \rightarrow (0,0)$ for $x = y^2$. From that I assumed I had to integrate in that "direction." However, that's not the case because $0 \le y \le 1$ are the values for $y$. Put another way $0 \le y \le 1 \Leftrightarrow 1 \ge y \ge 0$. Am I understanding? – Andrew Falanga Apr 20 '15 at 23:11
• Perhaps first it is easier to think of the integral without the arrows. If you do, then the lower limits are always the smaller values and the upper limits are the larger values (so that it doesn't matter which way you write the inequalities). The reason that the arrows are important is because Green's theorem assumes the curve is oriented "positively." This means that if you walk in the direction of the arrows, then the region described by your double integral is on the "left." You should be able to see this in your picture. – user197427 Apr 21 '15 at 0:12
• If the region is on the "right," then you would need to add a negative sign in front of the integral. – user197427 Apr 21 '15 at 0:12
• That explanation makes a lot of sense. Thank you. I think I've got it now. – Andrew Falanga Apr 21 '15 at 14:33