Understanding Symmetry in Double Integrals

[Linday12]

Homework Statement

Let D be the triangular domain given by 0$$\leq$$ y $$\leq$$3, (y/3)-1 $$\leq$$ 1-(y/3). Then


$$\int$$$$\int$$ (e-x$$^{5}$$e^(sqrt(1+y^2))

Homework Equations

The Attempt at a Solution

There is a quick way to solve it by breaking apart the double integral and then, apparently the x^5 part goes to 0, by symmetry? Anyways, I'm not sure why.

Then I'm left with the double integral of e, and since the domain is just an isosceles triangle, I can multiply the area of it by the e(constant) to get 3e.

So my question is, why does that other part go to zero? I can't visualize it. Knowing this would be a great help. Thank you! (And sorry about the latex, I couldn't quite work it out)

 

[LCKurtz]

Is the integral supposed to be

$$\iint_R e - x^5 e^{\sqrt{1+y^2}}dA$$

and is the region supposed to be

$$\frac y 3- 1 \le x \le 1 - \frac y 3$$

If so, yes, it is because of the x⁵. When you integrate an odd power of x over a symmetric interval [-a,a] you get 0. Set it up as a dx dy integral and you will see.

 

[Linday12]

Yes, that is it. I will make sure to do that. Thank you very much!