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$D\subset\mathbb{R}^{2}$ is an arbitrary bounded open set with $C^{1}$ boundary whose perimer P is finite. Let $f\,:\,\mathbb{R}^{2}\longrightarrow\mathbb{R}$ be a given $C^{1}$ function satisfying $|f(x,y)\leq1$ for all (x,y) in D. Trying to show $|\int\int_{D}\frac{\partial f}{\partial y}(x,y)dxdy\,|\leq P$

So this is of the exact form for green's theorem where f is really our P for Pi + Qj. the other term is 0 so $\frac{\partial Q}{\partial x}$ is 0 but does not imply Q = 0, so we still have a $\int Q\, dy$ term how do I find that?

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It's totally legit to apply Green's theorem to $f dx$ (i.e. take $Q = 0$). Then you just get: $$ \int_D \frac{\partial f}{\partial y} = \int_{\partial D} f$$Take absolute values, push the absolute value into the integral and apply the bound! – uncookedfalcon Nov 22 '12 at 5:56

From the comment, I'm saying consider: $$\left| \int_D \frac{\partial f}{\partial y}\right| = \left| \int_{\partial D} f \right| \leqslant \int_{\partial D} |f| \leqslant P \cdot \sup_{x \in \partial D} |f| \leqslant P $$

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yes that works if Q = 0 but I was saying you can't assume it's 0, all you know is it doesn't involve any x terms so there's another integral to add – elich Nov 22 '12 at 6:57
oh yea they would all have to be the same – elich Nov 22 '12 at 7:07

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