Question about double (and triple) integrals over a symmetric area

[farleyknight]

]This isn't a home work question in particular, but just want confirmation about a general idea.

So in Calc III, you have integrals of the form

$$\int_{-a}^a \int_{-\sqrt{a^2 - x^2}}^{\sqrt{a^2 - x^2}} x y dy dx$$which is the typical rectangular coordinates for a circle. Now, the integrand is the term x y, and you sum over the pieces in the circle, you'll have for each piece a negative and a positive part, which will sum to zero. And since this holds for all pieces, then the integral will be zero.My question is basically, what kind of rule of thumb can you use to check if the integral actually is zero versus some still mistake on your part? I would assume that if x and y are both odd powers of a polynomial then you'll always have these opposing pairs. Is that correct?

 

[lanedance]

you could do the integration.

with polar coordinates porbably wouldn't be too hard

otherwise this is a question about the symmetry of the function, or more correct in this case x & y are both 1D anti-symmetric functions

so in the 1D case
note f(x) = x is defined as an antisymmetric function as f(-x) = -f(x) so it is easy to show the integral is zero
$$\int_{-a}^{a}dx f(x) = \int_{-a}^{0}dx f(x)+ \int_0^{a}dx f(x)$$

now in the, use dummy variable x', then change variable (back to x) to x' = -x, dx = -dx'
$$\int_{-a}^{0}dx' f(x') = \int_{a}^{0}(-dx) f(-x) = - \int_{0}^{a}(-dx) (-f(x)) =- \int_{0}^{a}dx f(x)$$

giving the result
$$\int_{-a}^{a}dx f(x) = - \int_{0}^{a}dx f(x)+ \int_0^{a}dx f(x) = 0$$

 

[lanedance]

now knowing that what can you say about eh 2D case?