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This problem is being very difficult for me to solve, I need help.

Consider $F:\mathbb{R}^2\rightarrow\mathbb{R}$ of class $C^1$, suppose that the level curves of $F$ are closed and that $\nabla F$ is never $0$ for $x\neq0$. Consider the region $D$ between the curves $F=a$ and $F=b$. For each $r$ in $[a,b]$, let $c_r$ be the curve $F=r$. Let $f:D\rightarrow\mathbb{R}$ continuous.

I have to show that $$\int_Df=\int_a^b\bigg(\int_{c_r}\frac{f}{|\nabla F|}\bigg)dr$$

Usually when I ask something here I show my attempts or my observations, but in this case I couldn't even start doing this!

Thank you very much for helping!

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  • $\begingroup$ How can the level curves be closed, and yet $\nabla F$ is never $0$ with a $C^1$ function? If $\mathcal{C}$ is some closed level curve, then somewhere in its interior there is either a point of discontinuity, a point with undefined $\nabla$, or a point with $\nabla=0$. $\endgroup$
    – 2'5 9'2
    Apr 23, 2013 at 21:33
  • $\begingroup$ @alex.jordan Pretend that the graph of $F$ is a paraboloid; it doesn't matter if $\nabla F = 0$ at the origin as long as $0 < a < b$. I agree though that the question should be phrased so that $\nabla F \neq 0$ on the region $D$, however. $\endgroup$ Apr 23, 2013 at 21:35
  • $\begingroup$ @ABlumenthal But OP says that the domain is all of $\mathbb{R}^2$ and implies the given conditions apply for all of $F$'s domain. All this is stated before $D,a,b$ are mentioned. I guess it's clear now that you point it out that the $\nabla$ restriction should only apply to $D$. $\endgroup$
    – 2'5 9'2
    Apr 23, 2013 at 21:37
  • $\begingroup$ This problem comes from a book about vector analysis. I really dont know if there is some mistake there cause I just started to learn, but I think it's ok. The author is my teacher in he uses his book for years. If there was some mistake someone would tell him and he would tell us. $\endgroup$
    – Integral
    Apr 23, 2013 at 21:41
  • $\begingroup$ That is what I think. $\endgroup$
    – Integral
    Apr 23, 2013 at 21:42

1 Answer 1

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The formula that needs to be proven is basically the change of variables $(x,y)\rightarrow(F,\tau)$, where $F$ is the level of $f$ and $\tau$ is the natural parameter (curve length) on the level set. It is actually easier to go in the inverse direction: start from the expression on the right where we integrate the function $\frac{f}{|\nabla F|}$. Then, passing from $(F,\tau)$ to $(x,y)$, the integral will transform into $$ \int_D \frac{f}{|\nabla F|} \left|\frac{D(F,\tau)}{D(x,y)}\right|dxdy,$$ where $$ \left|\frac{D(F,\tau)}{D(x,y)}\right|=|\mathrm{det}\left(\begin{array}{cc} \frac{\partial F}{\partial x} & \frac{\partial F}{\partial y}\\ \frac{\partial \tau}{\partial x} & \frac{\partial \tau}{\partial y} \end{array}\right)|$$ is the jacobian of the variable change. Hence, all we have to show is that $$ |\nabla F|=\biggl|\left(\frac{\partial F}{\partial x}\frac{\partial \tau}{\partial y}-\frac{\partial F}{\partial y}\frac{\partial \tau}{\partial x}\right)\biggr|.\tag{1}$$ But, since the unit vector $\vec{n}$ orthogonal to $\nabla{F}$ (and hence tangent to the level curve) has the form $$ \vec{n}=\left(-\frac{1}{|\nabla F|}\frac{\partial F}{\partial y}, \frac{1}{|\nabla F|}\frac{\partial F}{\partial x}\right),$$ then $d\vec{l}=\vec{n}d\tau=(dx,dy)\Rightarrow d\tau=n_xdx+n_ydy$, so that $$ \frac{\partial\tau}{\partial x}=n_x=-\frac{1}{|\nabla F|}\frac{\partial F}{\partial y},\qquad \frac{\partial\tau}{\partial y}=n_y=\frac{1}{|\nabla F|}\frac{\partial F}{\partial x}.$$ Substituting these expressions into (1), we see that it is indeed satisfied.

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  • $\begingroup$ I dont understand how do you say $F$ is a variable...because $F$ is a function. I never saw this kind of treatment. $\endgroup$
    – Integral
    Apr 24, 2013 at 0:13
  • $\begingroup$ Imagine that your level sets are circles centered at the origin. Then $F$ and $\tau$ are a kind of radial and angular coordinate, respectively. $F$ is certainly a function of $x,y$ (just as $r=\sqrt{x^2+y^2}$ is), which is determined in a tricky way by $f$. $\endgroup$ Apr 24, 2013 at 0:21
  • $\begingroup$ Its beyond my mind...but thanks. $\endgroup$
    – Integral
    Apr 24, 2013 at 0:33
  • $\begingroup$ im just starting to learn this, Im incapable of thinking in $F$ as a "kind of radial something" and just put as a variable. I need something more basic and clear. $\endgroup$
    – Integral
    Apr 24, 2013 at 0:34
  • $\begingroup$ I can only re-suggest to work out an explicit example of what was called "cylindrical shells" above. Put simply, suppose that your function $f(x,y)$ depends only on the combination $x^2+y^2$. Then describe the level sets and try to relate the left and the right side of your relation in this special case. Then try to understand how the things will be modified if the level sets are more complicated - say, if your function depends on the combination $y+x^2$. Good luck. $\endgroup$ Apr 24, 2013 at 0:46

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