Double Integral Over a General Region

[pvortex]

Let [MATH]D[/MATH] be a region given as the set of [MATH](x,y)[/MATH] with [MATH]-\phi(x)\leq y\leq\phi(x)[/MATH] and [MATH]a\leq x\leq b[/MATH], where [MATH]\phi[/MATH] is a nonnegative continuous function on the interval [MATH][a,b][/MATH]. Let [MATH]f(x,y)[/MATH] be a function on [MATH]D[/MATH] such that [MATH]f(x,y)=-f(x,-y)[/MATH] for all [MATH](x,y)\in D[/MATH]. Argue that [MATH]\iint_D f(x,y) \,dA = 0 [/MATH].

I have [MATH]\int_{a}^{b}\int_{-\phi}^{\phi}f(x,y) dydx[/MATH], and I start by computing the inner integral first (in terms of y).
I split it into two regions: [MATH]\int_{-\phi}^{0}f(x,y)dy + \int_{0}^{\phi}f(x,y)dy[/MATH]I do the following substitutions for the first integral: [MATH]y\Rightarrow-u[/MATH], [MATH]dy\Rightarrow-du[/MATH], [MATH]-\phi\Rightarrow\phi[/MATH], and [MATH]0\Rightarrow0[/MATH]This leads me to the following computations: [MATH]\int_{\phi}^{0}-f(x,-u)du + \int_{0}^{\phi}f(x,y)dy = -\int_{0}^{\phi}f(x,u)du + \int_{0}^{\phi}f(x,y)dy[/MATH].
I have no idea what do from here. I assume I can just resubstitute [MATH]y\Rightarrow-u[/MATH] and [MATH]dy\Rightarrow-du[/MATH] or do a similar substitution, but I'm confused as to what happens with the upper bound of the first integral [MATH]\phi[/MATH] if I do so.

Any and all help is appreciated.

 

[HallsofIvy]

Your problem is that when you changed variables from y to u you did not change the limits of integration! The integral is \(\displaystyle \int_a^b \int_{-\phi(x)}^0 f(x, y)dydx\). Letting u= -y, u= -y, dy= -du and f(x,y) becomes f(x, -u)= -f(x, u). When y= 0, u= 0 and when \(\displaystyle y= -\phi(x)\), \(\displaystyle u= \phi(x)\). So \(\displaystyle \int_a^b \int_{-\phi(x)}^0 f(x, y)dydx= \int_a^b\int_{\phi(x)}^0 -f(x, u) (-du)\). The two negatives, in "-f(x,u)" and "-du" cancel but when we reverse the order or integration we get another: \(\displaystyle \int_a^b \int_{-\phi(x)}^0 f(x, y)dydx= -\int_a^b\int_0^{\phi(x)} f(x,u)du\) which then cancels the other integral to give 0.

 

[pvortex]

I did, in fact, change the limits, if you see what I wrote for my substitution.

[MATH]y\Rightarrow-u[/math], [math]dy\Rightarrow-du[/math], [math]-\phi\Rightarrow\phi[/math], and [math]0\Rightarrow0[/MATH]

I also got the final part you mentioned (you seem to be missing [MATH]dx[/MATH] yourself? It doesn't seem like you substituted [MATH]x[/MATH] anywhere, so it should've been [MATH]dudx[/MATH]. In any case, that's irrelevant since I was only working on the inner integral.)

What I don't understand is how [MATH]-\int_{a}^{b}\int_{0}^{\phi(x)}f(x,u)dudx[/MATH] cancels with [MATH]\int_{a}^{b}\int_{0}^{\phi(x)}f(x,y)dydx[/MATH] when it is in terms of u. Is is just the fact that [MATH]y=-u\Rightarrow u=y \Rightarrow y=-y[/MATH] because the integral is 0? That doesn't seem right to me.

 

[Steven G]

In the left integral you wrote just let y =u! (and dy =du). These are dummy variables so you can replace them with any letter you like and the results will not change.

 

[pvortex]

Thanks, I forgot about that property.

 

[Steven G]

Thanks, I forgot about that property.

You don't have to think of it as a property. You have no problem making more complicated substitution so why not make the substitution y=u?

 

[pvortex]

You don't have to think of it as a property. You have no problem making more complicated substitution so why not make the substitution y=u?

Not property in a strictly mathematical sense, which was a poor choice of words. I voiced my concern about making said substitution in the post above your initial one, and I was just fixated on that relationship