Double Integral - Weird Region

**utopiaNow** · April 5th 2009, 01:11 AM

The function I have is
$<br /> f(x, y) = c(y^2 - x^2) e^{-y}<br />$

with $|x| \leq y \leq (\frac{1}{4} - x^2)\ and -\alpha \leq x \leq \alpha$
where $\alpha \approx 0.207$

I want to integrate over this region. So I noticed its symmetric first of all. So my plan is to only consider the positive half first and then double the resulting area I find. Secondly I find that I can separate this region into 2 regions. (See attached image)
Region 1: A triangle with base = height = $\alpha$ .
Region 2: upper limit of y is arc created by $1/4 - x^2$ and lower limit is $\alpha$ . And x ranges from 0 to $\alpha$ .

I find when I go to do the double integral for region 2 I keep getting a negative answer.
The integral for region 2 I'm tyring is:
$<br /> \int_0^\alpha \! \int_\alpha^{\frac{1}{4} - x^2} f(x, y)\,dy\,dx.\<br />$

Any suggestions? Am I on the right track? Or completely off?

Thanks in advance.

**Moo** · April 5th 2009, 02:13 AM

Hello,

Yes, if you continue your reasoning, you may be able to get the correct answer.
But it could have been more straightforward (and what follows can be seen on your graph)

If you consider $0\leq x\leq \alpha$ (which is correct), then $x \leq y \leq \frac 14-x^2$
So the boundaries for y are :
$\max \{0,x\}$ , which is x, since $x \geq 0$
and $\frac 14-x^2$

The the integral is :
$\int_0^\alpha \int_{{\color{red}x}}^{1/4-x^2} f(x,y) ~dy ~dx$

And you can see on your graph, that if you care only of the right parts of R1 and R2, y is always > x, but always below 1/4 - x².
Does it look clear to you ?

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