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Let $D$ be a region given as the set of $(x, y)$ with $$a \leq x \leq b\quad\text{and}\quad-\Phi(x) \leq y \leq \Phi(x)$$ where $Φ$ is a nonnegative continuous function on the interval $[a, b]$.

Let $f(x, y)$ be a function on $D$ such that $$f(x, y) = - f(x, - y)$$ for all $(x , y) \in D$.

Argue that $$\displaystyle \iint_D f(x, y) dA = 0.$$

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Let $I$ the integral, make the substitution $t=-y$, which leaves $D$ invariant, to get that $I=-I$.

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Integrate with respect to $y$ first. $$\iint_D f(x,y)\,{\rm d}A=\int_a^b\int_{-\Phi(x)}^{\Phi(x)}f(x,y)\,{\rm d}y\,{\rm d}x$$ Use your condition on $f(x,y)$ to show that the inner integral is $=0$.

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let $I= \displaystyle \iint_D f(x,y)\,{d}A=\int_a^b\int_{-\Phi(x)}^{\Phi(x)}f(x,y)\,{d}y\,{d}x =\int_a^b\int_{-\Phi(x)}^{\Phi(x)}-f(x,-y)\,{d}y\,{d}x$, now replacing -y by t we get, $I = \displaystyle\int_a^b\int_{\Phi(x)}^{-\Phi(x)}f(x,t)\,{d}t\,{d}x = -\int_a^b\int_{-\Phi(x)}^{\Phi(x)}f(x,t)\,{d}t\,{d}x = -I$
now $I=-I$ thus $I = 0$.

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Careful! The inequalities $\Phi(x)\leq t\leq -\Phi(x)$ do not make sense, since $\Phi$ is a nonnegative function. – Per Manne Sep 28 '12 at 17:01

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